THE  LIBRARY 

OF 

THE  UNIVERSITY 

OF  CALIFORNIA 

SANTA  BARBARA 

COLLEGE 


PRESENTED  BY 

Mrs.    S.  0,  ?;erner 


■It  n  %^H,  x^'O  .^^.' «,, 


Digitized  by  tine  Internet  Arciiive 

in  2007  witii  funding  from 

IVIicrosoft  Corporation 


littp://www.arcliive.org/details/elementsofsolidgOObruciala 


ELEMENTS 


OF 


SOLID    GEOMETRY 


BY 

W.   H.   BRUCE,  A.M.,  Ph.D. 

PRESIDBNT   NORTH   T£XAS    STATE    NORMAL   COLLEGE 
ABTD 

C.    C.    CODY,  A.M.,  Ph.D. 

PROFESSOR   OF    MATHEMATICS,    SOUTHWESTERN    UNIVERSITY 


DALLAS,  TEXAS 

THE   SOUTHERN  PUBLISHING  COMPANY 

1914 


COPTEIGHT,   1912,   BT 

THE  SOUTHERN  PUBLISHING  COMPANY. 


CpA 
Q-S7 

B7 


UNTVERSiry  OP  CALlFUKMft 
lANTA  lARBARA  COLLEGE  UBRAKY 

f*  A  O  .O  (TN 
.>  6  ^:  t.  u 


CONTENTS 

BOOK  VI 

PAGX 

Definitions 1 

Lines  and  Planes  in  Space 10 

Dihedral  and  Polyhedral  Angles 16,  26 

BOOK  VII 

Definitions 31 

Polyhedrons 31 

Prisms 32 

Pyramids 46 

Regular  Polyhedrons 63 

Cylinders 65 

Cones 69 

BOOK  VIII 
The  Sphere 

Definitions 75 

Spherical  Angles 82 

Spherical  Polygons 83 

Measurement  of  Spherical  Surfaces 95 

Spherical  Volumes 102 


in 


INDEX  OF  DEFINITIONS 


SKCTlOIf 

Angle  of  lune 786 

Angle  of  spherical  polygon     .     763 

Axis  of  cone 701 

Axis  of  cylinder 682 

Axis  of  pyramid 639 

Bi-rectangular   spherical    tri- 
angle   775 

Cone 608-702 

Conical  surface 696 

Convex  polj'hedron  .  .  .  .  599 
Convex  spherical  polygon  .     ,     755 

Cube 611 

Cylinder     ......  678-682 

Cylindrical  surface  ....     677 

Diagonal  of  spherical  polygon  754 
Diameter  of  sphere  ....     724 

Dihedral  angle 568 

Directrix 677,  696 

Distance 532,  738 

Dodecahedron      600 

Edge  of  dihedral  angle  .  .  .  559 
Edge  of  polyhedral  angle  .  .  686 
Edge  of  polyhedron  ....    697 

Element 677,  696 

Equal  solids 613 

Face 659,  586,  597 

Face  angle 686 

Frustum  of  cone 711 

Frustum  of  pyramid      .     ,     .    643 

Generatrix 677,  696 

Great  circle  of  sphere  .  .  .  733 
Hexahedron 600 


SECTION 

Icosahedron 600 

Inclination 582 

Lateral  area  .  .  .  .  601,  638 
Lateral  edge  ....  601,  638 
Lateral  face  ....  601,638 
Lateral  surface     .     .     .      678,  698 

Lune 784 

Nappe 697 

Octahedron 600 

Parallelopiped 608 

Plane 522 

Plane  angle  of  dihedral  angle 

561,  664 

Polar  distance 740 

Polar  triangle 768 

Poles  of  circle 736 

Polyhedral  angle 586 

Polyhedron 597-599 

Prism 601-607 

Projecting  plane 680 

Projection 676 

Pyramid 638-643 

Radius  of  cylinder   ....     682 

Radius  of  sphere 723 

Regular  polyhedron ....  673 
Right  section  of  cylinder  .  .  687 
Right  section  of  prism  .     .     .     606 

Section 598 

Side  of  spherical  polygon  .     .     758 

Similar  cones 704 

Similar  cylinders 683 

Similar  polyhedrons ....    666 


IV 


INDEX   OF  DEFINITIONS 


BKCTIOK 

Slant  height    .     .     .   642, 644,  703 

Small  circle  of  sphere    .     .     .  734 

Sphere 722 

Spherical  angle 750 

Spherical  excess  of  polygon    .  800 

Spherical  excess  of  triangle    .  777 

Spherical  polygon     .     .     .     .  753 

Spherical  pyramid "  .     .     .     .  802 

Spherical  sector 804 

Spherical  segment    ....  806 

Symmetrical  polyhedral  angles  591 


SECTION 

Symmetrical  spherical  polygons  758 
Tangent .     .     .     684,  705,  727,  728 

Tetrahedron 600 

Trihedral  angle 589 

Tri-rectangular  spherical  tri- 
angle   776 

Vertex  .  .  .  586,  597,  638,  753 
Vertical  polyhedral  angles  .  592 
Vertical  spherical  polygons    .     757 

Volume 612 

Zone 786 


SYMBOLS  AND  ABBREVIATIONS. 


ax.     . 

.  axiom. 

0  . 

.  circle. 

cor.    . 

.  corollary. 

+  . 

.  plus. 

def.    . 

.  definition. 

—  . 

.  minus. 

iden. . 

.  identity. 

X    . 

.  multiplied  by. 

prop. . 

.  proposition. 

-,/,:. 

.  divided  by. 

post.  . 

.  postulate. 

.  is  equal  to  or  equivalent  to. 

cons. . 

.  construction. 

~   . 

.  is  similar  to. 

hyp.   . 

.  hypothesis. 

S  . 

.  is  congruent  to. 

rect.  . 

.  rectangle. 

>    . 

.  is  greater  than. 

rt. 

.  right. 

<    . 

.   .  is  less  than. 

8t. 

.  straight. 

•       ±   . 

.  is    perpendicular  to,   or  a 

^           • 

.  angle. 

perpendicular. 

A        . 

.  triangle. 

II     . 

.  is  parallel  to,  or  a  parallel. 

O 

.  parallelogram. 

Q.  E.  D.  (quod  erat  demonstrandum),  which  was  to  be  proved. 
Q.  E.  F.  (quod  erat  faciendum),  which  was  to  be  done. 

Note.    The  foregoing  are  used  also  in  the  plural,  as  =  means  •'  are 
equal  to,"  as  well  as  "is  equal  to." 


REFERENCES   TO   PLANE  GEOMETRY 

63.  The  sum  of  all  the  angles  about  a  point  is  equal  to  two 
straight  angles. 

65.  If  two  straight  lines  intersect,  the  vertical  angles  are 
equal. 

85.  In  congruent  figures  homologous  parts  are  equal. 

86.  Any  side  of  a  triangle  is  less  than  the  sum  of  the  other 
two,  and  greater  than  their  difference. 

91.  Two  triangles  are  congruent  if  they  have  two  sides  and 
the  included  angle  of  the  one  equal,  respectively,  to  two  sides 
and  the  included  angle  of  the  other. 

92.  Two  right  triangles  are  congruent  if  the  legs  of  the 
one  are  equal,  respectively,  to  the  legs  of  the  other. 

94.  Two  triangles  are  congruent  if  they  have  two  angles 
and  the  included  side  of  the  one  equal,  respectively,  to  two 
angles  and  the  included  side  of  the  other. 

95.  Two  right  triangles  are  congruent  if  a  leg  and  an  adja- 
cent acute  angle  of  the  one  are  equal,  respectively,  to  a  leg  and 
an  adjacent  acute  angle  of  the  other. 

108.  The  perpendicular  bisector  of  a  line  is  the  locus  of 
points  equidistant  from  the  extremities  of  the  line. 

109.  Two  points  each  equidistant  from  the  extremities  of  a 
line  determine  the  perpendicular  bisector  of  the  line. 

115.  Only  one  perpendicular  can  be  drawn  from  a  given  ex- 
ternal point  to  a  given  straight  line. 

116.  The  perpendicular  is  the  shortest  line  that  can  be 
drawn  from  a  given  point  to  a  given  line. 

117.  Two  oblique  lines  from  the  same  point  in  the  perpen- 
dicular to  a  given  line,  cutting  off  equal  segments  from  the 


viii  BOOK  VI.    SOLID  GEOMETRY. 

foot  of  the  perpendicular,  are  equal ;  and  of  two  lines  cutting 
off  unequal  segments  from  the  foot  of  the  perpendicular,  the 
one  cutting  off  the  greater  segment  is  the  greater  line. 

121.  Two  right  triangles  are  congruent  if  they  have  the  hy- 
potenuse and  an  acute  angle  of  the  one  equal,  respectively,  to 
the  hypotenuse  and  an  acute  angle  of  the  other. 

122.  Two  right  triangles  are  congruent  if  the  hypotenuse 
and  a  leg  of  the  one  are  equal,  respectively,  to  the  hypotenuse 
and  a  leg  of  the  other. 

123.  Two  triangles  are  congruent  if  they  have  the  three 
sides  of  the  one  equal,  respectively,  to  the  three  sides  of  the 
other. 

132.  If  two  triangles  have  two  sides  of  the  one  equal,  re- 
spectively, to  two  sides  of  the  other,  but  the  third  side  of  the 
first  greater  than  the  third  side  of  the  second,  the  angle  oppo- 
site the  third  side  of  the  first  is  greater  than  the  angle  opposite 
the  third  side  of  the  second. 

137.  Two  straight  lines  in  the  same  plane  perpendicular  to 
the  same  straight  line  are  parallel. 

139.  If  a  straight  line  is  perpendicular  to  one  of  two  par- 
allels, it  is  perpendicular  to  the  other. 

151.  Two  angles  having  their  right  sides  respectively  par- 
allel and  also  their  left  sides  parallel  are  equal,  whereas  if  the 
right  side  of  each  is  parallel  to  the  left  side  of  the  other  they 
are  supplementary. 

165.  The  sum  of  three  angles  of  a  triangle  is  equal  to  a 
straight  angle. 

166.  A  diagonal  divides  a  parallelogram  into  congruent  tri- 
angles. 

167.  The  opposite  sides  and  the  opposite  angles  of  a  paral- 
lelogram are  equal. 

169.   Parallels  included  between  parallels  are  equal. 


REFERENCES   TO   PLANE   GEOMETRY.  ix 

173.  If  two  sides  of  a  quadrilateral  are  equal  and  parallel, 
the  figure  is  a  parallelogram. 

176.  Two  parallelograms  are  congruent  if  two  sides  and  the 
included  angle  of  one  are  equal,  respectively,  to  two  sides  and 
the  included  angle  of  the  other. 

213.   Kadii  of  the  same  circle  or  of  equal  circles  are  equal. 

233.  In  the  same  circle,  or  in  equal  circles,  equal  arcs  are  sub- 
tended by  equal  chords  and  intercepted  by  equal  central  angles. 

234.  In  the  same  circle,  or  in  equal  circles,  equal  chords 
subtend  equal  central  angles  and  equal  arcs. 

240.  Ten  propositions,  in  all,  may  be  obtained  by  selecting 
any  two  of  the  following  conditions  for  the  hypothesis  and 
any  one  of  the  remaining  three  for  the  conclusion; 

1.  passes  through  the  center  of  the  circle. 

2.  bisects  the  chord. 

3.  is  perpendicular  to  the  chord. 

4.  bisects  the  minor  arc. 

5.  bisects  the  major  arc. 

246.  Through  three  points  not  in  the  same  straight  line  one 
circle,  and  only  one,  can  be  drawn. 

275.  If  two  variables  are  constantly  equal  and  each  ap- 
proaches a  limit,  their  limits  are  equal. 

277.  The  limit  of  the  product  of  a  constant  and  a  variable 
is  the  product  of  the  constant  by  the  limit  of  the  variable. 

283.  In  the  same  circle,  or  in  equal  circles,  central  angles 
have  the  same  ratio  as  their  intercepted  arcs. 

287.   A  central  angle  is  measured  by  its  intercepted  arc. 

330.  In  any  proportion  the  terms  are  in  proportion  by  alter- 
nation. 

331.  In  any  proportion  the  terms  are  in  proportion  by  in- 
version.   • 


A  straight  line  that 


X  BOOK  VI.    SOLID   GEOMETRY. 

338.  In  any  proportion  like  powers  of  the  terms  are  in  pro- 
portion, 

340.  A  line  parallel  to  one  side  of  a  triangle  divides  the 
other  sides  proportionally. 

343.  If  a  straight  line  parallel  to  the  side  BC  of.  a  triangle 
ABC  cuts  AB  at  D  and  AC  at  E,  then  DE  :BC  =  AD:  AB. 

363.  Similar  polygons  are  those  whose  homologous  angles 
are  equal  and  whose  homologous  sides  are  proportional. 

369.  Two  triangles  are  similar  if  their  sides  are  respectively 
proportional. 

370.  Two  triangles  are  similar  if  the  sides  of  the  one  are, 
respectively,  parallel  or  perpendicular  to  the  sides  of  the 
other. 

374.  The  homologous  altitudes  of  two  similar  triangles 
have  the  same  ratio  as  any  two  homologous  sides. 

376.  Two  similar  polygons  may  be  divided  into  the  same 
number  of  similar  triangles,  similarly  placed. 

391.  The  square  of  the  hypotenuse  of  a  right  triangle  is 
equal  to  the  sum  of  the  squares  of  the  two  legs. 

392.  The  square  of  either  leg  of  a  right  triangle  is  equal  to 
the  difference  of  the  square  of  the  hypotenuse  and  the  square 
of  the  other  leg. 

420.  The  area  of  a  parallelogram  is  equal  to  the  product  of 
its  base  and  altitude. 

421.  Parallelograms  having  equal  bases  and  equal  altitudes 
are  equivalent. 

425.  The  area  of  a  triangle  is  equal  to  half  the  product  of 
its  base  and  altitude. 

428.  Triangles  of  equal  altitudes  are  to  each  other  as  their 
bases. 


REFERENCES  TO  PLANE   GEOMETRY.  xi 

430.   Triangles  of  equal  bases  and  altitudes  are  equivalent. 

434.  The  areas  of  two  triangles,  having  an  angle  in  the  one 
equal  to  an  angle  in  the  other,  are  to  each  other  as  the  products 
of  the  sides  including  the  equal  angles. 

436.  The  areas  of  two  similar  polygons  are  to  each  other 
as  the  squares  of  any  two  homologous  sides. 

486.  If  the  number  of  sides  of  an  inscribed  polygon  be  in- 
definitely increased,  the  apothem  of  the  polygon  will  approach 
the  radius  of  the  circle  as  a  limit. 

488.  The  circle  is  the  limit  of  the  perimeters  of  regular  in- 
scribed and  circumscribed  polygons  and  the  area  of  the  circle 
is  the  limit  of  the  areas  of  these  polygons  when  the  number  of 
their  sides  is  indefinitely  increased. 


SOLID   GEOMETKY. 


Book  YI. 

Solid  geometry,  or  geometry  of  three  dimensions,  treats  of 
figures  whose  elements  are  not  all  in  the  same  plane.     (§  35.) 

522.  A  plane  is  a  surface  such  that  the  straight  line  joining 
any  two  of  its  points,  lies  wholly  in  the  surface.  (§  16.)  While 
regarded  as  indefinite  in  extent,  it  is  usually  represented  in 
diagrams,  by  parallelograms  that  lie  in  the  plane.  • 

A  plane  is  said  to  be  determined  by  points  or  lines,  when 
these  points  or  lines  fix  its  position  in  space. 

523.  A  plane  is  determined  by  any  three  given  points  which 
are  not  in  the  same  straight 

line. 

Two  points  determine  a 
line  (§  9),  but  do  not  deter- 
mine a  plane,  because  a 
plane  may  be  rotated  about 
any  given  line,  assuming,  in 
turn,  an  indefinite  number  of 
positions,  but  a  third  point, 
without  this  line,  fixes  the 
plane. 


2  BOOK  VI.     SOLID  GEOMETRY. 

524.  Because  two  points  determine  a  line  and  three  points 
a  plane,  so  a  plane  is  determined  by  the  equivalent  of  three 
points;  namely,  a  straight  line  and  a  point  without  this 
line;  two  intersecting  straight  lines;  two  parallel  straight 
lines. 

525.  The  point  in  which  a  straight  line  meets  a  plane  is 
called  the  foot  of  the  line. 

526.  A  straight  line  is  perpendicular  to  a  plane  when  it  is 
perpendicular  to  every  line  in  the  plane  drawn  through  its 
foot. 

527.  A  straight  line  is  parallel  to  a  plane,  or  a  plane  is 
parallel  to  a  straight  line,  when  the  two  will  not  meet  if  pro- 
duced indefinitely. 

528.  An  oblique  line  is  one  which  is  neither  perpendicular 
nor  parallel  to  a  plane. 

529.  Two  planes  are  parallel  if  they  will  not  meet  if  pro- 
duced indefinitely. 

530.  If  two  planes  are  not  paral- 
lel, they  must  intersect  in  a  line 
common  to  the  two  planes.  For  two 
planes  cannot  have,  in  common,  a 
straight  line  and  any  point  with- 
out that  line.  If  they  could,  the 
two  planes  would  coincide.  (§  623.) 
That  is,  the  common  points,  or  intersection,  cannot  be  other 
than  a  straight  line. 


BOOK  VI.    SOLID  GEOMETRY. 


Proposition  I.    Theobem. 


531.   The  perpendicular  is  the  shortest  line  that  can 
he  drawn  from  a  point  to  a  plane. 


Given  the  point  P,  the  line  PA  ±  the  plane  MN,  and  any  other 
line  PB  from  P  to  MN. 

To  prove  PA  <  PB. 

Proof.     Through  A,  the  foot  of  PA,  draw  the  line  AB. 

Then  PA  is  ±  AB.  §  626 

.-.  PA  <  PB.  §  116 

{The  JL  is  the  shortest  line  that  can  be  drawn  from  a  given 
point  to  a  given  line.)  Q.  e.  d. 

532.   The  distance  from  a  point  to  a  plane  is  the  length  of 
the  perpendicular  from  the  point  to  the  plane. 


BOOK  VI.     SOLID   GEOMETRY. 


Pboposition  II.     Theorem. 

533.  Oblique  lines  draicn  from  a  point  to  a  plane, 
meeting  the  plane  at  equal  distances  from  the  foot  of 
the  perpendicular,  are  equal ;  and  of  two  oblique  lines 
meeting  the  plane  at  unequal  distances  from  the  foot 
of  the  perpendicular,  the  more  remote  is  the  greater. 


Given  PO  JL  plane  MN,  and  AG  =  BG,  and  GD  >  GB. 
To  prove    I.  PA  =  PB. 

II.  PD  >  PB. 

Proof.     I.     The  rt.  As  POA  and  POB  are  congruent,  having 
PO  common  and  OA  =  OB.  Hyp.  §  92 

.■.PA  =  PB. 
II.     On  CD  take  OC  =  OB  and  draw  PO. 
Then  PD  >  PC. 

But  PC  =  PB. 

.:  PD  >  PB. 


§85 


§117 
Proof  I 

Q.  E.  D. 


634.  Cor.  Conversely  :  Equal  oblique  lines  from  a  point 
to  a  plane  meet  the  plane  at  equal  distances  from  tJie  foot  of  the 
perpendicular ;  and  of  two  unequal  lines  the  greater  meets  the 
plane  at  the  greater  distance  from  the  foot  of  the  perpendicular. 


BOOK  VI.     SOLID   GEOMETRY. 
Proposition  III.     Theorem. 


535.  A  straight  line  perpendicular  to  each  of  two 
strata^  lines  at  their  point  of  intersection  is  perpen- 
dicular to  the  plane  of  those  lines. 

A 


Given  AB  ±  BC  and  BD  at  B. 

To  prove  AB  1.  MN,  the  plane  of  these  lines. 
Proof.     Draw  BE,  any  other  line,  in  the  plane  MJ^. 
Draw  DC  meeting  BE  at  E,  and  prolong  AB  to  A'  so  that 
BA'  =  AB. 

Join  A  and  A'  with  D,  E,  and  C. 

BD  and  BC  are  each  ±.  AA'  at  its  mid-point. 

Hyp.  and  Cons. 
Hence  in  As  ADC  and  A'DC, 

AD  =  DA',  §  108 

also,  AC=A'C,  §108 

and  DC=DC.  Idem 

.-.  A  ADC  ^  A  A'DC.  §  123 

.-.  ^  ADE  =  ^  A'DE.  ^  85 

Hence  A  ADE  ^  A  A'DE,  §  91 

.-.  AE  =  EA\  and  ^iJ  ±  ^^'  at  B.  §§  109,  85 

•••  AB  1.  any  line  in  MN  passing  through  its  foot. 

.-.  it  is  J.  the  plane  MN.  §  526 

Q.  E.  D. 


6 


BOOK  VI.    SOLID  GEOMETRY. 


^v'^ 


Proposition  IV.     Theorem. 

536.  All  the  perpendiculars  to  a  straight  line  at  the 
same  point  lie  in  a  plane  perpendicular  to  that  line 
(Converse  of  §  535). 


Given  PB,  PC,  PD,  each  ±  line  AP  at  P. 
To  prove        PB,  PC,  PD,  lie  in  a  common  plane  X  AP. 
Proof.     Let  MN  be  the  plane  of  PB  and  PC.  §  624 

Then  ^P±  plane  JO:  §636 

Let  the  plane  of  AP  and  PD  intersect  the  plane  MN  in  the 

line  PD'.  §  530 

Then  AP  ±  PD\  §  626 

But  AP  ±  PD.  Hyp. 

Since  in  the  plane  APD  but  one  ±  can  be  drawn  to  AP  at 

P,  §116 

PD'  must  coincide  with  PD. 

.'.  PB,  PC,  and  PD  lie  in  a  common  plane  A.  AP. 

'  '  ^  Q.E.D. 

537.  Cor.  Tlirongh  a  given  point  hut  one  plane  can  he 
passed  perpendicular  to  a  given  line,  and  the  plane  which  is  the 
perpendicular  hisector  of  a  straight  line  is  the  locus  of  points  equi- 
distant from  the  extremities  of  the  line. 


BOOK  VI.     SOLID  GEOMETRY. 


Proposition  V.    Pkoblem. 

638.   Through  a  given  point,  to  draw  a  line  perpen- 
dicular  to  a  given  plane. 

I.   When  the  given  point  is  within  the  plane. 
II.   When  the  given  point  is  without  the  plane. 


I.   Given  point  P  within  the  plane  MN. 
Required  through  P  to  draw  a  _L  3/-^. 
Construction.     From  P  in  plane  MN  draw  any  line  PB. 
In  same  plane  draw  PD  JL  PB.  §  113 

Through  P  pass  a  plane  intersecting  MN  in  the  line  PJ5, 
and  in  this  plane  draw  PL  1.  PB. 

In  RS,  the  plane  of  PD  and  PL,  draw  AP  JL  PD.         §  113 

Then  PA  ±  MN. 

Proof.     Because  PB  is  ±  PD  and  PL,  it  is  1.  their  plane. 

§  535 

.■.PB±AP.  §526 

And  because  AP  is  X  PB  and  PD,  it  is  X  MN,  their  plane. 

§  535 

Ex.  556.    Find  the  locos  of  points  in  space  equidistant  from  all  points 
of  a  circle. 


8  BOOK  VI.     SOLID   GEOMETRY. 

II.   Given  point  P  without  the  plane  KN. 

Required  through  P  to  draw  a  l.to  MN. 

Construction.     Pass  through  P  a  plane  intersecting  MN  in 

some  line  BC,  and  in  this  plane  draw  BP  _L  BC  §  114 

In  the  plane  MN  draw  BD  ±  BC.  §  113 

And  in  the  plane  of  BD,  BP,  draw  PA  ±  BD.  §  114 

Then  is  PA  JL  MN. 

Proof.     The  A&  PAB,  ABC,  and  PBC  are  rt.  As.  Cons. 

.-.pa'  =  PB'-  A^.  §392 

AC'=AB'  +  B(f.  §391 

Pff  +  BC'  =  PC\  §391 

Adding,  PAl+A&=  PC\ 

.-.  AP^Cis  art.  A.  §391 
.-.  PA  ±  AC. 

..PA±MN.  Q.E.F. 

539.   Cor.     Through  a  given  point  but  one  perpendicular  to  a 
plane  can  be  drawn. 

Ex.  557.    Find  the  locus  of  a  point  in  space  equidistant  from  three 
given  points  not  in  the  same  straight  line. 

Ex.  558.     Find  a  point  equidistant  from  four  given  points  not  all  in 
the  same  plane. 


BOOK  VI.     SOLID   GEOMETRY.  9 

Proposition  VI.     Theorem. 

540.  If  from  the  foot  of  a  perpendicular  to  a  plane 
a  line  he  drawn  at  right  angles  to  any  line  of  the  plane, 
and  the  point  of  intersection  he  joined  ivith  any  point 
of  the  perpendicular,  the  last  line  will  he  perpendic- 
ular to  the  line  of  the  plane. 


Given  AF  ±  plane  MN,  FP  ±  any  line  CB  in  plane  MN,  and 
PA  drawn  from  P  to  any  point  of  AF. 


To  prove 

APA-BC. 

Proof. 

Erom  P  on  CB  take  PC=PB. 
Join  A  and  F  with  C  and  B. 

Then 

FC=FB. 

§108 

.'.AB  =  Aa 

§533 

Hence 

AP  ±  BC. 

§109 

Q.  E.  D. 

10 


BOOK  VI.     SOLID   GEOMETRY. 


V    /  Proposition  VII.    Theorem. 

541 .   Two  perpendiculars  to  the  same  plane  are  parallel. 


JE 


ID 


§113 
§540 


Given  AB  and  CD  ±  the  plane  MN. 
Toprcyoe  AB  W  CD. 

Proof.    Join  D  to  A  and  B.     Draw  EF  ±  BD. 
.•.AD±EF. 

Because  BD,  AD,  and  CD  are  each  J_  EF,  they  lie  in  same 
plane.  §  536 

AB  also  lies  in  this  plane.  Because  the  three  points  A,  B, 
and  D  determine  a  plane.  §  523 

.•.  AB  and  CD  lying  in  the  same  plane  and  _L  the  same  line, 
BD,  are  II.  §  137 

Q.  E.  D. 

^  iO  .A.         .     .,. 


\ ^N  ^ : *2»r 

542.   Cor.  1.     If  one  of  tioo  parallels  is  perpendicular  to  a 
plane,  the  other  is  perpendicular  to  the  same  plane. 

If  CD  is  not  ±MN,dvAyrCE±  MN.   Then  CEWAB.    §641. 
.-.  CE  and  CD  must  coincide.     .*.  CD  ±  MN. 


^k.  11 


BOOK  VI.     SOLID  GEOMETRY.  11 


543.  Cor.  2.  If  two  lines  are  II  a  third  line  they  are  II  each 
other. 

If  MN  is  drawn  ±  AB,  it  must  be  ±  CD  and  also  ±  EF. 
.'.  the  lines  are  II. 


Pboposition  VIII.     Theorem. 

544.  A  straight  line  parallel  to  a  line  in  a  plane  is 
parallel  to  the  plane. 


Given  AB  II  CD  in  plane  MN. 
To  prove  AB  II  plane  MN. 

Proof.    AB  and  CD  being  II  lie  in  the  same  plane  AD.  §  524 
.*.  if  AB  meets  MN,  it  must  meet  it  in  line  CD.      §  530 
But  AB  and  CD  are  II  and  cannot  meet. 

.•.  AB  II  plane  MN.  q.e.d. 

545.  Cor.  1.     If  a  line  is  parallel  to  a  plane,  the  intersection 
of  the  plane  with  any  plane  through  the  line  is  parallel  to  the  line. 

546.  Cor.  2.     If  two  intersecting  lines  are  each  parallel  to  a 
given  plane,  the  plane  of  these  lines  is  parallel  to  the  given  plane. 


12 


BOOK  VI.     SOLID   GEOMETRY." 


v(^ 


TWO  PLANES. 
Pbopositiox  IX.     Theorem. 


547.    Two  planes  perpendicular  to  same  straight  line 
are  parallel. 


M 


N 


-B 


Given  planes  MN  and  PQ  ±  AB. 

To  prove  the  plane  MN  II  the  plane  PQ. 

Proof.  If  MN  and  PQ  are  not  II,  they  will  meet  if  sufficiently 
produced. 

Suppose  them  to  meet.  We  would  then  have  two  planes 
through  the  same  point  _L  to  the  same  line,  which  is  impos- 
sible. §  537 

Therefore,  plane  MN  II  plane  PQ.  Q.  B.  d. 

548.  CoR.  If  a  straight  line  and  a  plane  are  perpendicular  to 
the  same  straight  line,  they  are  parallel. 


v^^ 


BOOK  VI.     SOLID  GEOMETRY. 


Proposition  X.     Theorem. 


13 


549.   The  intersections  of  two  j^cirallel  planes  hy  a 
third  plane  are  parallel  lines. 


Given  two  planes,  MN  and  PQ,  intersected  by  the  plane  RS  in 
AB  and  CD. 


To  prove 


AB  II  CD. 


Proof.     AB  and  CD  lie  in  the  same  plane  ^*S'.  Hyp. 

Because  they  also  lie  in  the  planes  MN  and  PQ  they  can- 
not meet.  §  529 


AB  II  CD. 


Q.  E.  D. 


550.  CoR.  1.     Parallel  lines  included  between  parallel  planes 
are  equal. 

551.  Cor.  2.     Parallel  planes  are  everywhere  equally  distant. 


14 


BOOK    VI.     SOLID  GEOMETRY. 


vW 


Proposition  XI.     Theorem. 


552.  A  straight   line  perpendicular   to  one  of  two 
parallel  planes  is  perpendicular  to  the  other,  also. 


Given  MN  and  PQ,  II  planes,  and  AB  ±  plane  PQ. 
To  prove  AB  ±  plane  MN. 

Proof.  Draw  BE  and  BF,  any  two  lines  in  PQ  intersecting 
atB. 

Suppose  planes  passed  through  AB-BE  and  AB-BF,  inter- 
secting MN  in  AC  and  AD,  respectively. 

Then  AC  II  BE  and  AD  II  BF.  §  649 

But  ABl.  BE  Qxid.  to  BF.  §526 

.-.  AB  A.  AC  and  to  AD.  §  139 


.*.  ABl. plane  MN. 


§535 

Q.  E.  D. 


553.  CoR.  1.     Reciprocally,  a  plane  perpendicular  to  one  of 
two  parallel  lines  is  perpendicular  to  the  other,  also. 

554.  Cor.  2.     Throiigh  a  given  point  one  plane,  and  only  one, 
can  he  drawn  parallel  to  a  given  plane. 


BOOK  VI.     SOLID  GEOMETRY. 
Proposition  XII.     Theorem. 


16 


555.  Two  angles  not  m  the  same  plane,  having  their 
sides  respectively  parallel,  right  side  to  right  side  and 
left  side  to  left  side,  are  equal  and  their  planes  are 
parallel. 


Given  ^s  o  and  o'  lying  in  planes  MN  and  PQ,  respectively,  and 
having  sides  AB  II  A'B',  AC  II  A'C. 

To  prove  "^  o  =  "if  o'  and  MN  II  PQ. 

Proof.     Take  A'B'  =  AB  and  A'C  =  AO.    Draw  AA',  BB', 
CC. 
Because  AB  and  AC  are  respectively  II  and  =  A'B'  and  A'C, 


Hence 


But 


Also 


AA'B'B  and  AA'CC  are  Os. 

§173 

.  AA'  =  and  II  BB'  and  AA'  =  and  II  CC 

§167 

BB'  =  and  II  CC. 

§543 

.'.  CB'  is  a  O,  and  BC  =  B'C. 

§167 

.'.  A  ABC  ^  A  A'B'C. 

§123 

.-.  ^  0  =  ^  o\ 

§85 

^o'  =  ^  o". 

§65 

.-.  ^  0  =  ^  o". 

PQ  is  II  AB  and  AC. 

§545 

,'.MN\\PQ. 

§546 

Q.  B.  D. 

16 


BOOK  VI.     SOLID   GEOMETRY. 


556.  Cob.  Two  angles  not  in  the  same  plane,  having  their 
sides  respectively  parallel,  right  side  to  left  side  and  left  side  to 
right  side,  are  supplementary  and  their  planes  are  parallel. 

Propositiox  XIII.     Theorem. 
557.   If  two  straight  lines  are  intersected  hy parallel 
planes,  the  corresponding  segments  are  proportional. 

N 


\  X- 

t       \ 

M 

\       " 

\  L- 

N      \             S 

\         „ 

„ '^J^    \ 

\      ^ 

^     \ 

EB     FD  GD 


Given  AB  and  CD,  intersected  by  planes  MN,  PQ,  and  RS,  in 
A,  E,  B,  and  C,  G,  D,  respectively. 

To  prove  4E  =  9E. 

'  EB     GD 

Proof.  Draw  AD  intersecting  PQ  in  2?:  Draw  EF,  BD,  FG,  AC. 

Then  EF  II  BD  and  FG  II  AC.  §  549 

:^  §340 

FD 

.  ^  =  9^ 
"  EB     GD' 

Q.E.D. 
DIHEDRAL  ANGLES. 

558.  When  two  planes  intersect,  their  divergence  from  the 
line  of  intersection  is  called  a  dihedral  angle. 

559.  The  two  planes  are  its  faces  and  the  line  of  intersection 
is  its  edge. 


BOOK  VI.     SOLID   GEOMETRY. 


17 


A   b 


B 

^le  inter- 

e    B 


560.  When  a  dihedral  angle  stands  alone  it  may  be  desig- 
nated by  the  two  letters  on  its  edge,  as 
the  dihedral  angle  AB.  When  two  or 
more  dihedral  angles  have  a  common 
edge,  each  angle  is  designated  by  four 
letters,  as  angles  C-AB-D,  or  D-AB-E. 

561.  The  plane  angle  of  a  dihedral  angle  is  the  angle  formed 

by  two  lines,  one  in  each  face,  drawn  q 

perpendicular  to  the  edge  at  the  same 
point.  Thus  ^  ohc  and  i^  def  are  plane 
angles  of  the  dihedral  ^  C-AB-D.  ^  ^ 

A  plane  perpendicular  to  the  edge  of  a  dihedral  angle  inter- 
sects the  faces  in  lines  which  form  the 
plane  angle  of  the  dihedral. 

562.  Any  two  plane  angles  of  a  dihe- 
dral angle  or  of  equal  dihedral  angles 

^      *         ^  ahc  =  ^  def,  their  sides  being  II.  §  555 

Since  the  points  b  and  e  are  taken  anywhere  on  the  edge, 

the  plane  angle  of  a  dihedral  angle  is  of  the  same  magnitude  at 

every  point  of  the  edge. 

Conversely,  two  dihedral   angles  are  equal   if  their  plane 

angles  are  equal. 

Given  the  plane  ^  BAC  =  ^  DEF.     The  edge  AH  A.  the  plane 

BAC  and  EK  ±  plane  DEF.     Then  by 

superposition   it   may  be  shown  that 

the    dihedral    ^    B-HA-C  =  dihedral 

^  D-KE-F.  §  524 

563.  A  dihedral  angle  may  be  con- 
ceived as  generated  by  the  revolution 
of  a  plane  about  a  line  taken  as  an 
edge.  As  the  magnitude  of  the  plane 
angle  depends  upon  the  amount  of  the  revolution  of  one  of  its 
sides,  from  an  initial  line,  independent  of  the  length  of  its 


K 


B 


E 


!  i 


D 


18 


BOOK  VI.     SOLID  GEOMETRY. 


sides  (§  57),  so  the  magnitude 
of  a  dihedral  angle  depends  upon 
the  amount  of  the  revolution  of 
a  plane  from  an  initial  plane, 
and  is  independent  of  the  ex- 
tent of  the  planes. 

564.  By  passing  a  plane  perpendicular  to  the  edge  of  the 
dihedral  angle  it  is  evident  that  the  plane  angle  is  the  measure 
of  the  dihedral  angle.  A  dihedral 
angle  is  acute,  right,  obtuse,  straight, 
or  reflex,  according  as  its  plane 
angle  is  acute,  right,  obtuse,  straight, 
or  reflex. 

565.  Dihedral  angles  are  adja- 
cent, vertical,  complementary,  or 
supplementary  just  as  their  plane  angles  hold  these  relations. 

566.  Many  of  the  theorems  of  lines  and  angles  have  analo- 
gous theorems  with  planes  and  dihedral  angles.  By  forming 
right  sections  of  the  dihedral  angles  the  proofs  of  the  follow- 
ing theorems  may  be  obtained  from  the  corresponding  theorems 
of  plane  geometry : 

1.  The  supplements  of  equal  dihedral  angles  are  equal. 

2.  The  complements  of  equal  dihedral  angles  are  equal. 

3.  Vertical  dihedral  angles  are  equal. 

4.  If  two  parallel  planes  are  cut  by  a  third  plane,  the  alternate 
interior  dihedral  angles  are  equal ;  the  interior  dihedral  angles 
on  the  same  side  of  the  transverse  plane  are  supplementary ; 
the  corresponding  dihedral  angles  are  equal ;  and  conversely. 

5.  Two  dihedral  angles  whose  faces  are  respectively  parallel 
are  either  equal  or  supplementary. 

6.  Two  dihedral  angles  whose  edges  are  parallel  and  whose 
faces  are  respectively  perpendicular  are  either  equal  or  supple- 
mentary. 


tv^^ 


BOOK  VI.     SOLID  GEOMETRY. 


Proposition-  XIY.    Theorem. 


19 


567.  If  a  straight  line  is  perpendicular  to  a  plane^ 
every  plane  containing  that  line  is  perpendicular  to  the 
plane. 


Given  AB  ±  plane  MN,  and  PQ  any  plane  passing  through  AB 
intersecting  MN  in  CQ. 

To  prove  plane  PQ  ±  plane  MN. 

Proof.     In  Jf^draw  BD  ±  CQ.  §  113 

But  AB  ±  CQ.  §  526 

.'.  ^  ABD  is  the  plane  ^  of  the  dihedral  ^  P-QC-N.  §  561 

And  ^  ABD  is  a  rt.  ^.  §  526 

.-.  plane  PQ  ± plane  MK  §  564 

Q.E.D. 

568.   Cor.     A  plane  perpendicular  to  the  edge  of  a  dihedral 
angle  is  perpendicular  to  its  faces. 


Proposition  XV.    Theorem. 


569.  If  tivo  planes  are  perpendicular  to  each  other, 
any  straight  line  in  one  plane  drawn  2^erpendicular  to 
their  intersection  is  perpendicular  to  the  other. 


20 


BOOK  VI.     SOLID   GEOMETRY. 

Pr 

A 


D. 


M^ 


C- 


^B 


lE 


Given  plane  PQ  ±  plane  MN,  and  AB  in  PQ  ±  DE,  their  in. 
tersection. 

To  prove  AB  ±  plane  MN. 

Proof.     In  plane  MN  draw  BC  ±  DE.  §  113 

Then  ^  ABC  is  the  plane  :^  of  the  dihedral  ^  P-DE-M. 

§561 
But  dihedral  ^  P-DE-M  is  a  rt.  dihedral  ^.  Hyp. 

.'.  ^  -4£C,  its  measure,  is  a  rt.  ^.  §  564 

Also  AB  1.  DE.  Hyp. 

.•.  AB,  being  ±  ^O  and  Z)^  at  their  intersection,  is  1.  their 

plane.  §  535 

.•  AB  _L  plane  MN.  Q.  e.  d. 

570.  Cor.  1.  If  tivo  planes  are  perpendicular  to  each  other, 
and  a  perpendicular  to  one  of  them  is  drawn  from  any  point  in 
their  intersection,  it  will  lie  in  the  other  plane.  . 

571.  Cor.  2.  If  two  planes  are  peipendicular  to  each  other, 
and  a  perpend icidar  to  one  of  them  is  drawn  from  any  point  in 
the  other  plane,  it  icill  lie  in  that  plane. 


/\V>\^ 


Proposition  XVI.     Theorem. 


572.  If  two  intersecting  planes  are  each  perpendicu- 
lar to  a  third  plane,  the  line  of  intersection  is  perpen- 
dicular to  that  2)lane. 


BOOK  VI.     SOLID   GEOMETRY. 


21 


Given  planes  PQ  and  RS,  each  ±  plane  MN,  and  intersecting 
in  AB. 

To  prove  AB  X  plane  MN. 

Proof.     At  B  in  plane  MN  erect  a  ±  to  MN. 

This  ±  lies  in  each  of  the  planes  RS  and  PQ.  §  570 

.".it  coincides  with  AB,  their  intersection.  §  530 

.  • .  AB  A.  plane  MN.  q.  e  d 

573.  Cob.  Conversely,  if  a  plane  is  perpendicular  to  each  of 
two  intersecting  planes,  it  is  perpendictdar  to  their  line  of  inter- 
section. 

Proposition  XVII.     Theorem. 

574.  The  plane  bisecting  a  dihedral  angle  is  the  locus 
of  points  equidistant  from  its  faces. 


22  BOOK  VI.     SOLID  GEOMETRY. 

I.  Given  the  dihedral  "^  A-BG-C,  the  plane  BD  bisecting  this 
"i^,  any  point  P  in  the  plane  BD,  PE,  and  PF,  ±s  drawn  from  P 
to  the  faces  AB  and  BC. 

To  prove  Pis  equidistant  from  the  faces  AB  and  BC,  or 

PE=PF. 

Proof.  1.  Through  PE  and  PF  pass  a  plane  intersecting 
AB  in  EG,  BC  in  GK,  and  BD  in  PG. 

Because  PE  ±  AB  and  PF  ±  BC,  Cons, 

the  plane  PEF  ±  BG,  their  intersection.  §  573 

.-.  BG  ±  EG,  PG,  and  FG.  §  535 

.-.  ^s  EGP  and  PGF  are  the  plane  ^s  of  the  dihedral  ^s 

A-BG-D  and  D-BG-C.  §  561 

But  the  ^  A-BG-D  =  ^  D-BG-C.  Hyp. 

.-.  ^  EGP=  ^  PGF.  §  562 

.-.  rt.  A  PEG  ^  rt.  A  PEG.  §  121 

.'.  PE  =  PF.  §85 

II.  Given  P'  any  point  without  the  bisecting  plane  BD. 

To  prove  P'  unequally  distant  from  the  faces  AB  and  AC. 

Proof.  Draw  PK±  BC,  P'E  ±  AB,  and  from  P  draw  PF  ± 
BC.  Pass  a  plane  through  P'E  and  P'K  intersecting  AB  in 
EG,  BD  in  PG,  and  BC  in  GK. 

Then     PP  +  PF>PF.     (§86.)     Also  P'2i^>P'^.     §531 

.-.  PP'  +  PF>  P'K  and  PF=  PE.  Proof  I 

.-.  PP  +  PE>  P'K  or  PE  >  P'/iT. 

.-.  any  point  without  the  bisecting  line  is  unequally  distant 
from  the  faces  of  the  dihedral  angle.  Q.  e.  d. 

675.  Dep.  The  projection  of  a  point  on  a  plane  is  the  foot  of 
the  perpendicular  from  the  point  to  the  plane. 


BOOK  VI.     SOLID  GEOMETRY. 


23 


The  projection  of  a  line  on  a  plane  is  the  locus  of  the  projec- 
tion of  its  points  on  the  plane. 

This  may  be  illustrated  by  a  shadow  cast  by  a  ruler  on  a 
wall,  when  the  rays  of  light  casting  the  shadow  fall  perpen- 
dicularly upon  the  wall. 

Proposition  XVIII.    Problem. 

576.  Through  a  straight  line  not  perpendicular  to  a 
given  plane  to  pass  a  plane  perpendicular  to  that  plane. 


Given  the  line  AB  not  ±  plane  MN. 

Required  to  pass  a  plane  through  AB  1.  MN. 

Ccnstruction.     From  P,  any  point  in  AB,  draw  PK  X  plane 


MN. 


§  538 


Through  PK  and  AB  pass  a  plane  AF. 
AF  is  the  plane  required. 
Proof.     The  plane  AF  through  AB  intersects  MN. 
.'.  plane  AF  ±  plane  MN. 


Cons. 
§567 

Q.  E.  F. 

677.  Cor.  1.  Tlirough  a  straight  line  not  perpendicular  to  a 
given  plane  only  one  plane  can  he  passed  perpendicular  to  that 
plane. 

578.  Cor.  2.  The  projection  of  a  straight  line,  not  perpen- 
dicular to  a  plane,  upon  that  plane,  is  a  straight  line. 


24 


BOOK  VI.     SOLID  GEOMETRY. 


579.  Cor.  3.  The  projection  of  a  straight  line,  perpendicular 
to  a  plane,  is  a  point. 

580.  Def.  The  plane  containing  all  the  perpendiculars 
drawn  from  a  straight  line  to  a  plane  is  called  the  projecting 
plane. 

Proposition  XIX.     Theorem. 

581.  The  acute  angle  ivhich  a  line  makes  with  its  pro- 
jection on  a  plane  is  the  least  angle  which  it  makes  with 
any  line  of  that  plane. 


Proof. 


Given  BC  the  projection  of  the  line  AB  on  the  plane  MN,  and 
BD  any  other  line,  drawn  in  plane  MN,  through  B. 

To  prom  -^  ABC  <  ^  ABD. 

Take  BD=BC,  draw  AD  and  AC. 

lu  the  As  ABC  and  ABD, 

AB  =  AB. 

BD  =  BC. 

AC  <:AD. 

.:  :^  ABC <^  ABD. 


But 


Iden. 
Cons. 
§531 


§132 

Q.  B.  D. 

582.    Def.     The  inclination  of  a  line  to  a  plane  is  the  acute 
angle  which  the  line  makes  with  its  projection  on  that  plane. 


BOOK   VI.     SOLID  GEOMETRY. 


25 


Proposition  XX.     Problem, 

583.   To  draw  a  common  perpendicular  to  tioo  lines 
not  in  the  same  plane. 

A  K  B 


AT 

\    \     '^- 

A 

\            E                     XP 

A 

\                                          ^^D                      \ 
m\ X 

Given  two  lines  AB  and  CD  not  in  the  same  plane. 
Required  to  draw  a  common  _L  to  AB  and  CD. 
Construction.     Through  any  point  P  of  CD  draw  EF  II  AB. 

Through  EF  and  CD  pass  a  plane  MN. 
Then  MNisWAB.  §544 

Through  AB  pass  a  plane  AH  1.  MN  (§  576),  intersecting 
MN  in  GH.  Then  GH  II  ^i^  (§  545),  and  must  intersect  CD 
in  some  point  L. 

At  L  in  the  plane  AH  draw  i/iTX  G^.  §  113 

Then  LK  is  the  ±  required. 

Proof.     Since  G^IT  II  AB,  and  iif  ±  GH  (cons.),  i^  X  ^5. 

§  139 

But  LKl.  MN.  §  569 

.-.  LKl.  CD.  §  526 

Q.  E.  F. 

584.  Cor.  1.  Only  one  common  perpendicular  can  he  drawn 
to  two  lines  not  in  the  same  plane. 

585.  Cor.  2.  The  common  perpendicular  is  the  shortest  path 
between  two  lines  not  in  the  same  plane. 


26 


BOOK  VI.     SOLID  GEOMETRY. 


POLYHEDRAL   ANGLES. 

586.   When  three  or  more  planes  meet  at  a  common  point  the 
angle  thus  formed  is  called  a  polyhedral  angle. 

y 

V 


The  common  point  is  the  vertex  of  the  angle,  the  intersections 
of  the  planes  are  the  edges,  the  portions  of  the  planes  between 
the  edges  are  the  faces,  and  the  plane  angles  formed  by  the 
edges  at  the  vertex  are  the  face  angles. 

Thus  in  the  diagram,  V  is  the  vertex,  VA,  VB,  VC,  etc.,  are 
the  edges,  VAB,  VBC,  etc.,  are  the  faces,  and  the  angles  AVB, 
BVC,  etc.,  are  face  angles. 

587.  A  polyhedral  angle  is  designated  by  the  letter  at  the  ver- 
tex, as  the  polyhedral  angle  F;  or  by  the  letter  at  the  vertex  and 
a  letter  at  a  point  on  each  edge,  as  the  polyhedral  angle  V-ABC. 

588.  A  polyhedral  angle  whose  base  is  a  convex  polygon  is 
called  a  convex  polyhedral  angle. 

589.  A  polyhedral  angle  of  three  faces  is  called  a  trihedral 
angle. 


BOOK  VI.     SOLID   GEOMETRY. 


27 


590.  Two  polyhedral  angles  are  congruent  when  their  face 
angles  and  their  dihedral  angles  are  equal,  each  to  each,  and  are 
arranged  in  the  same  order,  for  they  may  be  made  to  coincide. 

591.  Two  polyhedral  angles  are  S3nnmetrical  when  their  face 
angles  and  their  dihedral  angles  are  equal  each  to  each  but  ar- 
ranged in  a  reverse  order. 

V 


Two  symmetrical  polyhedral  angles  cannot,  generally,  be 
made  to  coincide ;  hence  to  show  their  equivalence,  an  indirect 
method  is  necessary. 

592.  Two  polyhedral  angles  are  vertical  if  the  edges  of  one 
are  the  prolongations  through  the  vertex  of  the  edges  of  the 
other. 

Proposition  XXI.     Theorem. 

593.  The  sum  of  any  two  face  angles  of  a  trihedral 
angle  is  greater  than  the  third  face  angle. 


28 


BOOK  VI.     SOLID  GEOMETRY. 


Given  the  trihedral  '^  V,  with  ^  AVC  >  :^  AVB  or  ^  BVC. 
To  prove  ^  AVB  +  ^BVC>^AVa 
Proof.     In  ^  AVC  draw  VD,  making  ^AVD  =  :^AVB. 
Through  D,  any  point  in  VD,  draw  AC. 
Take  F5  =  VD,  and  draw  ^B  and  BC. 
Then  A^F^^A^FD.  §91 

.'.AB=AD.  §85 

In  A  ^BC,  AB  +  BC>  AC.  §  86 

But  AB  =  AD. 

By  subtraction,  BC  >  (AC -AD)  or  J5C >  DC. 
In  As  5FC  and  DVC,  VC  is  common, 

VB  =  VD,  Cons. 

and  BC  >  DC.  Just  proved 

.•.:^BVC>^DVC.  §132 

But  ^  ^  FB  =  ^  ^  FD.  Cons. 


Adding,     ^  ^FB  +  ^  £F(7  >  ^  ^FD  +  ^  ZJFC. 
.-.  ^  y1FB+  ^  5FC'  >  ^  ylFC. 

Proposition  XXII.     Theorem. 


Q.  B.  D. 


594.    71ie  sum  of  the  face  angles  of  any  convex  pohj' 
hedral  angle  is  less  than  tvjo  straight  angles. 


BOOK  VI.     SOLID  GEOMETRY. 


29 


Given  the  polyhedral  ^  V. 

To  j)rove  the  sum  of  the  ^s  at  V  less  than  2  st.  ^s. 

Proof.  Pass  a  plane  cutting  the  polyhedral  ^  in  the  polygon 
ABODE. 

From  0,  any  point  in  this  polygon,  draw  OA,  OB,  00,  OD,  OE. 

The  number  of  ^s  having  a  common  vertex  0  is  the  same  as 
the  number  of  2^s  having  a  common  vertex  V. 

.'.  the  sum  of  all  the  ^s  in  the  As  having  a  vertex  at  0  = 
the  sum  of  all  the  ^s  of  the  As  having  a  vertex  at  V.        §  155 

But  ^  VBA  +  ^  VBO  >  :^  ABO, 

and  :f  VOB  +  ^  VOD  >  ^  BOD,  etc.  §  593 

.'.  the  sum  of  the  base  ^s  of  the  As  having  a  common  vertex 
at  Fis  >  the  sum  of  the  base  ^s  of  the  As  having  a  common 
vertex  at  0. 

.'.  the  sum  of  the  ^s  at  the  vertex  at  F<  the  sum  of  the  ^s 
at  the  vertex  0. 

But  the  sum  of  the  angles  at  the  vertex  0  =  2  st.  ^s.       §  63 

.'.  the  sum  of  the  l^s  at  V  <2  st.  ^s.  Q.  e.  d. 

Proposition  XXIII.     Theorem. 

595.  If  two  trihedral  angles  have  the  three  face  angles 
of  one  equal  to  the  three  face  angles  of  the  other,  they 
are  either  congruent  or  symmetrical. 


30  BOOK  VI.     SOLID   GEOMETRY. 

Given  the  trihedral  ^s  V  and  V,  having  the  face  ^s  AVB,  AVC, 
and  BVC  =  the  face  ^s  A'V'B',  A'V'C,  and  B'V'C,  respectively: 

To  prove  the  corresponding  dihedral  ^s  of  V  and  V  are  =, 
and  that  the  trihedral  ^s  V  and  V  are  congruent  or  symmetrical. 

Proof.     On  the  edges  of  V  and  V  take  VA,  VB,  VC,  V'A', 

V'B',  V'C  all  equal,  and  draw  AB,  BC,  AC,  A'B',  B'C,  A'C. 

Then  As  VAB,  VBC,  VAC  ^  As  V'A'B',  V'B'C,  V'A'C.  §  91 

.-.  A  ABC  ^  A  A'B'C.  §  123 

In  the  edges  VC,  V'C,  take  VD=  V'D'  and  in  the  faces 
VBC,  VAC,  and  V'B'C,  V'A'C,  draw  DE,  DF,  and  D'E', 
D'F'  X  the  edges  VC  and  V'C  respectively,  meeting  BC,  AC, 
B'C,  and  A'C  in  E,  F,  E',  and  F".    Draw  EF,  E'F'. 

Then  A  CED  ^  A  C'E'D'.  §  95 

.■.CE=CE', 

and  DE  =  D'E'. 

Likewise  CF=  CF', 

and  DF=  D'F'. 

.'.A  CFE  ^  A  CF'E'.  §  91 

.■.FE  =  F'E'.  §85 

.-.  A  FED  ^  A  F'E'D'.  §  123 

.•.:^FED  =  -^  F'E'D'.  §  85 

.-.  dihedral  ^  VC=  dihedral  ^  V'C. 

In  like  manner  it  can  be  proved  that  dihedral  ^  VB  =  dihe- 
dral ^  V'B'  and  dihedral  ^  VA  =  dihedral  ^  VA'.  §  562 

.'.  the  trihedrals  V  and  V'  are  congruent  or  symmetrical 
according  as  the  equal  face  angles  are  arranged  in  the  same  or 
reverse  order.  q.e.  d. 

596.  Cor.  If  two  trihedral  angles  have  three  face  angles  oftlie 
one  equal  to  the  three  face  angles  of  the  other,  then  the  dihedral 
angles  of  the  one  are  respectively  equal  to  the  dihedral  angles  of  the 
other. 


Book  YII. 
polyhedrons. 

597.  A  polyhedron  is  a  solid  TDounded  by  planes.  These 
planes  are  the  faces;  the  intersections  of  the  faces,  the  edges, 
and  the  intersections  of  the  edges,  the  vertices  of  the  polyhe- 
dron. 

A  line  joining  any  two  vertices,  not  in  the  same  face,  is  a 
diagonal. 

A  polyhedron  cannot  have  less  than  four  faces. 

598.  A  section  of  a  polyhedron  is  the  figure  formed  by  its 
intersection  with  a  plane. 

599.  A  polyhedron  is  convex  when  every  section  of  it,  formed 
by  the  intersection  of  a  plane  with  its  faces,  is  a  convex  polygon. 

All  polyhedrons  considered  in  this  book  are  convex. 

600.  A  polyhedron  of  four  faces  is  called  a  tetrahedron ;  of 
six  faces,  a  hexahedron ;  of  eight  faces,  an  octahedron  ;  of  twelve 
faces,  a  dodecahedron  ;  of  twenty  faces,  an  icosahedron. 


Tetrahedron.  Hexahedron.  Octahedron. 

31 


32 


BOOK  VII.     SOLID  GEOMETRY. 


Dodecahedron. 


Icosahedron. 


Prism. 


PRISMS   AND  PARALLELOPIPEDS. 

601 .  A  prism  is  a  polyhedron,  of  which  two  faces  are  congruent 
polygons  in  parallel  planes,  and  the  other  faces  are  parallelo- 
grams. The  congruent  polygons  are  the  bases 
and  the  parallelograms  are  the  lateral  faces  of 
the  prism.  The  intersections  of  the  lateral  faces 
are  the  lateral  edges.  The  sum  of  the  areas 
of  the  lateral  faces  is  the  lateral  area.  The  per- 
pendicular distance  between  its  bases  is  the 
altitude.  The  lateral  edges  of  a  prism  are  all 
equal,  because  parallel  lines  included  between 
parallel  planes  are  equal. 

602.  A  prism  whose  lateral  edges  are  "perpendicular  to  its 
bases  is  a  right  prism. 

Hence,  the  lateral  edge  of  a  right  prism  is  equal  to  its  alti- 
tude. 

603.  A  right  prism  whose  bases  are 
regular  polygons  is  a  regular  prism. 

604.  A  prism  whose  lateral  edges  are 
not  perpendicular  to  its  bases  is  an 
oblique  prism. 

605.  A  prism  is  triangular,  quadran- 
gular, etc.,  according  as  the  bases  are 
triangles,  quadrilaterals,  etc. 


POLYHEDRONS. 


33 


606.   The  section  of  a  prism  made  by  a  plane  perpendicular 
to  its  lateral  edges  is  a  right  section. 


607.  A  truncated  prism  is  that  part  of  a  prism  included 
between  a  base  and  a  section  made  by  a  plane 

oblique  to  the  base. 

608.  A  parallelepiped  is  a  prism  whose  bases  are 
parallelograms.  Hence  all  the  faces  of  a  paral- 
lelopiped  are  parallelograms. 

609.  A  right  parallelepiped  is  a  parallelopiped 
whose  lateral  edges  are  perpendicular  to  its 
bases. 


610.  A  rectangular  parallelopiped  is  a  right  parallelopiped 
whose  bases  are  rectangles. 

Hence  all  its  faces  are  rectangles. 

611.  A  cube  is  a  parallelopiped  whose  faces  are  squares. 
Hence  all  of  its  edges  are  equal. 

612.  The  volume  of  any  solid  is  the  ratio  of  the  solid  to 
another  solid  taken  arbitrarily  as  the  unit  of 

volume.  Because  of  its  convenience  a  cube 
whose  edge  is  a  linear  unit  is  adopted  as  the 
unit  of  volume. 


613.   Equal  solids  are  solids  whose  volumes 
are  equal. 


34  BOOK  VII.     SOLID  GEOMETRY. 

Proposition  I.     Theorem. 

614.    Sections  of  a  prism  made  hy  parallel  planes 
cutting  all  the  lateral  edges  are  congruent  polygons. 


Given  the  prism  PM  cut  by  II  planes  making  the  sections  AD 
and  A'D'. 

To  prove  section  AD  ^  section  A'D'. 

Proof.     AB,   BC,    CD,  etc.  II  A'B',  B'C,  CD',  etc.,  respec- 
tively. §  549 
.-.  AB,  BC,  CD,  etc.  =  A'B',  B'C,  CD',  etc.,  respectively. 

§169 

Also  ^s  ABC,  BCD,  etc.  =  ^s  A'B'C,  B'C'D',  etc.,  respec- 
tively. §  655 
.-.  polygon  AD  ^  polygon  A'D'. 

(for  the  polygons  have  their  sides  and  ^s  equal  each  to  each,  and  are 
therefore  congruent) .  Q.  B.  D. 

615.  CoR.  Any  section  of  a  prism  made  by  a  plane  parallel 
to  the  base  is  equal  to  the  base;  also  all  right  sections  of  a  prism 
are  equal. 


POLYHEDRONS, 


35 


Proposition  II.     Theorem. 


616.  The  lateral  area  of  a  prism  is  equal  to  the 
product  of  the  perimeter  of  a  right  section  by  a  lateral 
edge. 


Given  the  prism  PM.    Denote  its  lateral  area  by  s,  a  lateral  edge 
by  e,  and  the  perimeter  of  a  right  section  by  p. 

To  prove  s=p  xe. 

Proof.  PF=KG  =  LH,  etc.  =e.         §  601 

Also  AB  ±  KG,  BC±LH,  etc.  §606 

.-.  area  O  FK=  AB  x  PF,  =ABxe, 

area  CJ  GL  =  BCxe, 

and  area  O  ^.V  =  CD  x  e,  etc.  §  420 

But  the  sum  of  these  Os  equals  the  lateral  area  a. 

.-.  s  ={AB  +  BC  +  CD  +  etc.)x  e. 

And  the  sum  of  AB  ■i-BC-\-CD  +  etc.  =  p. 

.-.  s=p  X  e.  Q.E.D. 

617.  Cor.    The  lateral  area  of  a  right  prism  equals  the  prod- 
uct of  the  perimeter  of  the  base  by  the  altitude. 


86 


BOOK  VII.     SOLID  GEOMETRY. 
Proposition  III.    Theorem. 


618.  Tivo  prisms  are  congruent  when  the  three  faces, 
including  a  trihedral  angle  of  the  one,  are  congruent,  re- 
spectively, to  the  three  faces,  including  a  trihedral  angle 
of  the  other,  and  are  similarly  placed. 


Given  the  prisms,  AK  and  A'K',  having  the  faces  AD,  AL, 
AG  =  respectively  to  faces  A'D',  AX',  A'G',  and  similarly  placed. 

To  prove  AK^A'K'. 

Proof.  The  face  '^s  BAE,  EAF,  BAF  are  equal  respectively 
to  the  face  ^s  B'A'E',  E'A'F',  B'A'F.  Hyp. 

.-.  trihedral  ^  ^  =  trihedral  ^  ^'.  §  595 

Apply  the  prism  FD'  to  the  prism  FD  so  that  trihedral 
"^  A'  will  fall  on  its  congruent  trihedral  ^  A,  and  the  faces  of 
trihedral  ^  A'  shall  coincide  i^ith  the  congruent  faces  of  tri- 
hedral ^  A,  and  the  points  C,  D  shall  fall  on  the  points  C",  D'. 

Since  the  points  L\  F,  G'  coincide  with  L,  F,  G,  the  planes 
F'K'  and  FK  coincide  (§  523),  because  the  lateral  edges  of 
the  prism  are  II. 

.-.the  edges  C'ff,  D'K'  coincide  with  CH,  DK,  and  the 
points  i?',  iT' coincide  with  IT, /iT.    .'.  AK^A'K'.  q.e.d. 


POLYHEDEONS. 


37 


619.  Cor.  1.  Two  truncated  prisms  are  congruent  when  the 
three  faces  including  a  trihedral  angle  of  the  one  are  congruent  to 
the  three  faces  including  a  trihedral  angle  of  the  other. 

620.  Cor.  2.  Two  prisms  with  equivalent  bases  and  equal 
altitudes  are  equal. 

Proposition  IV.     Theorem. 

621.  An  oblique  prism  is  equivalent  to  a  right  prism 
whose  hase  is  a  right  section  of  the  oblique  j^rism  mid 
whose  altitude  is  a  lateral  edge  of  the  oblique  prism. 


B  G 

Given  the  oblique  prism  PM  with  the  right  section  A'M'  and 
lateral  edge  PA ;  also  the  right  prism  P'M'  with  base  A'M'  and 
lateral  edges  each  equal  to  PA. 

To  prove  PM=PW. 

Proof.  PA'  =  PA. 

Subtracting  PA'  from  each  of  these  equals, 
P'P  =  A' A. 
R'R  =  B'B. 
PR  =  AB  and  P'R'  =  A'B'. 
2(:s  of  PR'  =  ^s  of  AB>. 
.-.  face  PR'  ^  face  AB'. 
face  RS'  ^  face  BC 


Likewise 

Also 

And 


Hyp. 


§167 
§151 
§176 


Similarly, 


88 


BOOK  VII.     SOLID  GEOMETRY. 


And  base  PN=  AM.  §  601 

.-.  truncated  prism  PN'  =  truncated  prism  AM',  §  619 
Hence                 PM  -FN'=  PM  -  AM', 

or  PM=P'M'.  Q.E.D. 

Proposition  Y.     Theorem. 

622.   The  opposite  lateral  faces  of  a  parallelopiped 
are  parallel  and  congruent. 


H_ 


E 


/rr~ 


X 


a 


Given  parallelepiped  AG. 
To  p)'ove 

n  AFWand  ^O DG,  and  n AH  Wand^O BG, 

Proof.     Because  AC  is  a  O,  §  608 

.-.  AB  II  and  =  DG,  §  167 

But  AH"  is  a  O,  §608 

.'.AE\\Q.udi=DH.  §167 

.-.  ^  EAB  =  ^  HBGy 

and  plane  AF II  plane  DG,  §  656 

.-.OAF^ODG.  §176 

Similarly,       O  AH  II  and  ^  O  ^G'.  Q.  e.  d. 

623.   Cor.     Any  two  opposite  faces  of  a  parallelopiped  may  be 
taken  as  bases. 


POLYHEDRONS. 


89 


Pboposition  VI.     Theorem. 


624.  The  plane  passed  through  two  diagonally  oppo- 
site edges  of  a  parallelopiped  divides  it  into  two 
equivalent  triangular  prisms. 


Given  the  plane  QSBC  passing  through  the  diagonally  opposite 
edges  QC  and  SB  of  the  parallelopiped  PD. 

To  prove  PD  is  divided  by  the  plane  into  two  equivalent  tri- 
angular prisms,  ACB-S  and  BCD-B. 

Proof.     Let  EFGH  be  a  right  section  of  PD  intersected  by 
the  plane  QSB  C  in  the  diagonal  IIF. 


Then 


EF II  HG  and  EH  II  FG. 

.:  EFGH  is  a  O. 

.-.  AEFH^AHFG. 


§549 
§164 
§166 


The  triangular  prism  ACB-S  =  a  right  prism  whose  base 
is  EHF  and  altitude  AP,  and  the  prism  BCD-B  =  a  right 
prism  whose  base  is  FHG  and  altitude  AP. 

But  these  right  prisms  are  equal,  having  equal  bases  and 

equal  altitudes.  §  621 

.-.  A CB-P=  BCD-B.  Q.  E.  D. 


40 


BOOK  VU.     SOLID  GEOMETRY. 


Pboposition  VII.     Theorem. 


625.   Tiuo   rectangular  parallelopipeds   having   con 
gruent  bases  are  to  each  other  as  their  altitudes. 


FT" 

"    \ 

\ 

^N 

D_ 


Given  two  rectangular  parallelopipeds  P  and  Q  having  congruent 
bases  and  the  altitudes  AB  and  CD. 


To  prove 


Q     CD' 


Case  I.     When  AB  and  CD  are  commensurable. 

Proof.     Let  AKhe  the  common  measure  of  AB  and  CD. 
Suppose  it  is  contained  m  times  in  AB  and  n  times  in  CD. 


Then 


AB^m 
CD      n 


At  the  points  of  division  of  AB  and  CD  pass  planes  II  to 
the  bases. 

These  planes  divide  P  into  m  and  Q  into  n  equal  parallelo- 
pipeds. §  620 

"Q      n      ^     Q      CD' 


POLYHEDRONS. 


41 


Case  II.    When  AB  and  CD  are  incommensurable. 

Bwr r, V 


K 


Proof.  Divide  AB  into  any  number  of  equal  parts  and 
apply  one  of  these  parts,  as  AK,  to  CD  as  a  unit  of  measure. 
Since  AB  and  CD  are  incommensurable,  there  will  be  a  re- 
mainder ED  less  than  one  of  the  parts. 

Through  E  pass  a  plane  II  the  bases  of  Q  and  let  Q'  be  the 
parallelepiped  between  this  plane  and  the  lower  base  of  Q. 


Then 


q^^CE 
P     AB 


Case  I 


If  the  unit  of  measure  for  AB  be  continually  decreased,  the 
remainder  ED,  which  is  always  less  than  the  unit  of  measure, 
may  be  made  smaller  than  any  assignable  quantity,  but  not 
equal  to  zero,  since  AB  and  CD  are  incommensurable. 

.*.  CE  will  approach  CD  as  a  limit. 

.'.  ■ will  approach as  a  limit. 

AB  ^^  AB 

0'  0 

.'.  -^  will  approach  ^  as  a  limit. 


§275 
§331 

Q.  £  D 


p 

_CD 
~  AB 

p 
Q 

AB 
CD 

42 


BOOK  Vn.     SOLID  GEOMETRY. 


626.  Def.     The  three  edges  of  a  rectangular  parallelepiped 
meeting  at  a  common  vertex  are  called  its  dimensions. 

627.  Cor.     Two  rectangular  parallelopipeds  which  have  two 
dimensions  in  cominon  are  to  each  other  as  their  third  dimension. 

Pboposition  VIII.     Theorem. 

628.  Two  rectangular  parallelopipeds  having  equal 
altitudes  are  to  each  other  as  their  bases. 


/' 

/ 

/^ 

/  " 

a 

A 

V 

oJ  ■  ■■■ 

a 

c 

Given  two  rectangular  parallelopipeds  P  and  Q,  having  a  com- 
mon altitude  a  and  the  dimensions  of  their  bases  b,  c,  and  b',  c', 
respectively. 

Q 


To  prove 


b'Xc' 

Proof.     Construct  a  third  rectangular  parallelepiped  M,  hav- 
ing the  same  altitude  a  and  the  dimensions  of  the  base  c  and  b'. 
Then  Q  and  M  have  two  dimensions,  a  and  b'  in  common. 

.-.-  =  -.  §627 

Q     c< 

Likewise  P  and  R  have  two  dimensions  a  and  c  in  common. 


.  P  _ 

"  B~ 

b 

Multiplying 

the 

equations, 

P 

b-xc 

Q 

b'x& 

627 


Q.B.D. 


POLYHEDRONS, 


43 


629.  CoK.  Two  rectangular  parallelopipeds  having  one  di- 
mension in  common  are  to  each  other  as  the  products  of  the  other 
two  dimensions. 

Proposition  IX.     Theorem. 

630.  Tivo  rectangular  parallelopipeds  are  to  each 
other  as  the  products  of  their  three  dimensions. 


/ 

/ 

/  / 

x^ 

/ 

/-/ 

p 

a 

,       y 

R 

a 

h 

/c 

b 

yc' 

Given  the  two  rectangular  parallelopipeds  P  and  Q  having  the 
dimensions  a,  b,  c  and  a',  b',  c',  respectively. 


To  prove 


P _   axbxc 
Q~a'  X  b'  xc'' 


Proof.     Construct  a  third  rectangular  Darallelopiped  R  having 
the  dimensions  a,  b,  and  c'. 


Then 

id 

B 

Q 
p_ 

R 

axb 
a'x6' 

c 
''c'' 

Multiplying 

the 

equations, 

P 

axb 

xc 

§  629 
§  627 


Q     a'xb'xc' 


Q.E.D. 


44 


BOOK  VII.     SOLID   GEOMETRY. 


Proposition  X.    Theobem. 


631.    The  volume  of  a  rectangular  parallelopiped  is 
equal  to  the  product  of  its  three  dimensions. 


Given  any  rectangular  parallelopiped  P,  with  dimensions  a,  b, 
c,  and  the  cube  U,  the  unit  of  volume,  whose  edge  is  the  linear 
unit. 


To  prove 

P=a xbx  c 

Proof. 

P     axbxc 
U     1x1x1 

§  630 


p  . 

Because  TJ  is  the  unit  of  volume,  —  is  the  numerical  meas- 
ure of  the  volume  P.  ^  §  612 


the  volume  of  P  =  a  x  6  X  c. 


Q.E.  D. 


632.  Cor.  1.  TJie  volume  of  a  cube  is  equal  to  the  cube  of 
its  edge. 

633.  CoR.  2.  Tlie  volume  of  a  rectangular  parallelopiped  is 
equal  to  the  product  of  its  base  by  its  altitude. 

NoTK.  By  the  statement  of  Proposition  X  is  meant  that  the  number 
of  unit  cubes  in  the  volume  of  any  rectangular  parallelopiped  is  equal  to 
the  product  of  the  numerical  measures  of  its  length,  breadth,  and 
thickness, 


f 


POLYHEDRONS. 


Proposition  XI.    Theorem. 


45 


634.    The  volume  of  any  parallelopiped  is  equal  to 
the  product  of  its  base  hy  its  altitude. 


Given  the  oblique  parallelepiped  P,  with  base  B  and  altitude  H. 

To  prove  volume  P=Bx  H. 

Proof.  Produce  the  edge  AO  and  the  edges  II  it.  On  AO 
produced  take  EF  =  AC  and  through  E  and  F  pass  planes  JL 
the  produced  edges,  forming  the  oblique  parallelopiped  Q  whose 
base  B'  is  a  rectangle. 

Produce  the  edge  FG  and  the  edges  II  it.  Take  KL  =  FG 
and  through  L  and  K  pass  planes  _L  these  edges,  forming 
the  rectangular  parallelopiped  M. 

Then  P=Qa,ndQ  =  R  §  621 

Also  B  =  B' SindB'  =  B' .  §421 

Because  the  planes  of  the  upper  and  lower  bases  are  11  the 
three  parallelopipeds  have  a  common  altitude  H. 

But  volume  R  =  B"  x  H.  §  633 

.-.  volume  Q  =  B'  X  H, 

and  volume  P=B  x  H.  q.e.d. 


46 


BOOK  VII.     SOLID   GEOMETRY. 


Pboposition  XII.     Theorem. 


635.    TJie  volume  of  a  triangular  prism  is  equal  to 
the  product  of  its  hase  and  altitude. 


Given  the  triangular  prism  ADC-M  having  its  base  B  and 
altitude  H. 

To  prove  volume  ADC-M  =  BxH. 

Proof.  Upon  DA,  DC,  DM  as  edges  construct  the  paral- 
lelopiped  ADCR-M. 

Then    volume  ADC-M  =  |  volume  of  DS.  §  624 

But  volume  DS  =  ADCR  x  H=  2BxII.  §  634 

.-.  volume  ADC-M  =  ^x2BxH=BxH.  q.e.d. 

636.  Cor.  1.  TJie  volume  of  any  prism  is  equal  to  the  prod- 
uct of  its  hase  by  its  altitude. 

Any  prism  may  be  divided  by  diagonal 
planes  into  triangular  prisms.  These  prisms 
have  a  common  altitude.  The  volume  of 
each  is  equal  to  the  base  multiplied  by  the 
common  altitude.  Hence  the  sum  of  the  vol- 
umes of  the  triangular  prisms  is  equal  to  the 
sum  of  the  bases  multiplied  by  the  common 
altitude  H.  The  sum  of  the  triangular  prisms 
is  equal  to  the  given  prism  and  the  sum  of  the 
bases  is  equal  to  B,      .    y—  BxH 


POLYHEDRONS. 


47 


637.  Cor.  2.  Prisms  that  have  equivalent  bases  and  equal 
altitudes  are  equal;  jjrisms  are  to  each  other  as  the  product  of 
their  bases  by  their  altitudes;  prisms  having  equivalent  bases  are  to 
each  other  as  their  altitudes;  prisms  having  equal  altitudes  are  to 
each  other  as  their  bases. 

THE  PYRAMID. 

638.  A  pyramid  is  a  polyhedron,  one  of  whose  faces,  called 
the  base,  is  a  polygon,  and  the  lateral  faces  are 
triangles  having  a  common  vertex. 

The  lateral  edges  are  the  intersections  of  the 
lateral  faces. 

The  altitude  of  a  pyramid  is  the  perpendicu- 
lar from  the  vertex  to  the  plane  of  the  base. 

The  lateral  area  is  the  sum  of  the  areas  of  the 
lateral  faces. 

639.  When  the  base  is  a  regular  polygon,  the  center  of 
which  coincides  with  the  foot  of  the  altitude,  the  pyramid 
is  regular  and  its  altitude  is  its  axis. 

640.  A  pyramid  is  triangular,  quadrangular,  pentagonal, 
etc.,  as  its  base  is  a  triangle,  quadrilateral,  pentagon,  etc. 

641.  A  triangular  pyramid  is  also  called  a  tetrahedron  be- 
cause it  has  four  faces. 

642.  The  slant  height  of  a  regular  pyra- 
mid is  the  altitude  of  any  one  of  its  lateral 
faces. 

643.  A  truncated  pyramid  is  the  part  of  a 
pyramid  contained  between  the  base  and  a 
section  made  by  a  plane  cutting  all  its  lat- 
eral edges.  If  the  plane  of  the  section  is 
parallel  to  the  base,  the  part  between  this  section  and  the  base 
is  called  a  frustum  of  a  pyramid. 


48 


BOOK  VII.     SOLID  GEOMETRY. 


644.  The  altitude  of  a  frustum  of  a  pyramid  is  the  perpen- 
dicular between  the  planes  of  its  bases.  The  lateral  faces  of 
a  frustum  of  a  regular  pyramid  are  congruent  isosceles  trape- 
zoids and  its  slant  height  is  the  altitude  of  any  lateral  face. 

645.  The  lateral  edges  of  a  regular  pyramid  are  equal,  and 
the  lateral  faces  of  a  regular  pyramid  are  equal. 


Pbopositiox  XIII.     Theorem. 

646.  The  lateral  area  of  a  regular  'pyramid  is  equal 
to  half  the  product  of  the  slant  height  by  the  perimeter 
of  the  base. 


Given  A-BCDEF,  a  regular  pyramid,  L  its  slant  height,  P  the 
perimeter  of  its  base,  and  S  its  lateral  area. 

To  prove  S  =  ^  L  X  P. 

Proof.     The  lateral  faces  ABC,  ACD,  etc.,  are  equal  isoa- 
celes  A.  §  645 

The  common  slant  height  L  is  the  altitude  of  each  A. 
.'.  the  area  of  each  lateral  face  is  =  ^L  x  its  base.         §  425 
.'.  the  sum  of  all  the  lateral  faces  =^  L  X  sum  of  bases. 


But  the  sum  of  the  bases  =  P. 


.•.S  =  hLxP. 


Q.  E.  D. 


POLYHEDRONS.  49 

647.  Cob.     The  lateral  area  of  the  frus-  ^^-^>>«.^ 
turn  of  a  regular  pyramid  is  equal  to  half  the  AT      \     y\ 
product  of  the  slant  height  multiplied  by  the  /  /  ]      \    \     \ 
sum  of  the  perimeters   of  its   bases.     If  P  L-h']""       V"""-) 
is  the  perimeter  of  its  lower  base  and  p  is  \    /              \   / 

the  perimeter  of  its  upper  base,  then  S  =  ^  ^ ^ 

(P+P)L. 

Ex.  559.  Find  the  locus  of  si  point  in  space  equidistant  from  two 
given  intersecting  lines. 

Ex.  560.  The  planes  bisecting  the  dihedral  angles  of  a  trihedral  angle 
intersect  in  the  same  straight  line. 

Ex.  561.    The  lateral  faces  of  a  right  prism  are  rectangles. 

Ex.  562.    The  diagonals  of  a  parallelopiped  bisect  one  another. 

Ex.  663.    The  diagonals  of  a  rectangular  parallelopiped  are  equal. 

Ex.  564.     The  square  of  a  diagonal  of  a  rectangular  parallelopiped 

is  equal  to  the  sum  of  the  squares  of  its  three  dimensions. 

Ex.  565.  The  sum  of  the  squares  of  the  diagonals  of  a  parallelopiped 
is  equal  to  the  sum  of  the  squares  of  its  twelve  edges. 

Ex.  566.  Two  rectangular  parallelopipeds  with  equal  altitudes  have 
the  dimensions  of  their  bases  5  and  12,  and  9  and  20,  respectively.  What 
is  the  ratio  of  their  volumes  ? 

Ex.  567.    Find  the  volume  and  the  area  of  the  surface  of  a  cube  whose 

edge  is  12  ft. 

Ex.  568.  Find  the  edge  of  a  cube  equal  in  volume  to  a  rectangular 
parallelopiped  whose  dimensions  are  6  ft.,  8  ft.,  and  36  ft.,  respectively. 

Proposition  XIV.    Theorem. 

648.  If  a  pyramid  is  cut  hy  a  plane  parallel  to  the 
hose, 

I.    The  edges  and  altitude  are  divided  proportionally. 
II.    The  section  is  a  polygon  similar  to  the  base. 


60 


BOOK  VII. 

SOLID 

GEOMETRY. 

L 

y 

/ 

i  7 

m-jlM 

t    /" 

/ 

\ 

/'<7 

■>'    1 

/        ^ 

( 

^                   / 

Given  the  P3rramid  V-ABCDE  cut  by  the  plane  QR  II  the  base 
and  intersecting  the  lateral  edges  in  a,  b,  c,  d,  e  and  the  altitude  in  p. 

Toprove  I.    Z«  =Z^  =  J^=  ...  ^Yp.. 

VA     VB      VC  VP 

II.   The  section  abode  ~  base  ABODE. 
Proof.     I.  Through  vertex  V  pass  a  plane  II  QR. 
Then  the  edges  VA,  VB,  VC,  etc.,  and  the  altitude  VP,  being 
intersected  by  parallel  planes,  are  cut  proportionally.        §  657 

"VA     VB     VC      '"       VP' 

II.  Since  ah  II  AB,  he  II  BC,  cd  II  CD,  etc.  (§  549),  and  ^  abc 
=  ^  ABC,  '^hcd  =  :^  BCD,  etc.  (§  555),  then  the  two  polygons 
abcde  and  ABCDE  are  mutually  equiangular. 

Since^=i^,    ^=n.,,.    ^=_^.  §343 

AB     VB'    BC      VB         AB     BG 

Similarly,  • — -  =  — — ,  etc. 

•^'  BO     CD' 

Hence  the  homologous  sides  of  the  polygons  are  propor- 
tional. 

.-.  section  ahcde  ~  base  ABCDE.  §  360 

Q.  E.  D. 

Ex.  569.  How  many  bricks,  each  8  in.  Jong,  4  in.  wide,  and  2  in. 
thick,  are  equal  in  volume  to  a  wall  16  ft,  long,  4  ft.  wide,  and  12  ft. 
high? 


POLYHEDRONS. 


61 


649.   Cor.  1.     Parallel  sections  of  a  pyramid  are  to  each  other 
as  the  squares  of  their  distances  from  the  veiiex. 


Tor 
But 


abode     _  ab 
ab  _Vp 


=  --S-  or 


ab 


Vp 


AB      VP 


VP 


§436 
§338 


dbcde 


Vp_, 

' '  ABODE     vP 

650.  Cor.  2.  If  two  pyramids  having  equal  altitudes  are  cut 
hy  planes  imrallel  to  their  bases,  and  at  equal  distances  from  their 
vertices,  the  sections  have  the  same  ratio  as  their  bases. 


For 
But 


abode    -jy^also--^ 


so 


ABODE     vP"  ^^^     so" 

VP=SosindVP=SO. 


dbcde 


abode     ABODE 


§649 

Hyp. 
§330 


651.  Cor.  3.  If  two  pyramids  have  equal  altitudes  and  equiv- 
alent bases,  sections  made  by  planes  parallel  to  the  bases,  and  at 
equal  distances  from  the  vertices,  are  equivalent. 


52 


BOOK  VII.     SOLID  GEOMETRY. 


Proposition  XV.     Theorem. 


652.   Two   triangular  pyramids   having  equal  altir 
tildes  and  equivalent  bases  are  equivalent. 


Given  two  triangular  pyramids  P-ABC  and  P'-A'B'C',  with 
equivalent  bases,  ABC  and  A'B'C',  and  the  same  altitude  H. 

To  prove  P-ABC  =  P-A'B' C. 

Proof.  So  place  the  pyramids  that  their  bases  shall  be  in 
the  same  plane. 

Divide  the  altitude  H  into  n  equal  parts,  and  through  the 
points  of  division  pass  planes  II  to  the  bases. 

The  corresponding  sections  of  the  pyramids  will  be  equiv- 
alent. §  651 

Upon  these  sections  as  upper  bases  inscribe  a  series  of 
prisms,  a,  h,  c,  etc.,  a',  h',  c',  etc.,  in  each  pyramid. 

.-.  a  =  a',  b  =  b',  and  each  prism  in  P  =  corresponding  prism 
in  I"  and  a  -{-  b  +  c,  etc.,  =  a'  +  &'  +  c',  etc. 

Let  the  number  of  equal  parts  into  which  H  is  divided  be 
increased  indefinitely,  then  the  sum  a  +  b  +  c  +  etc.,  will  ap- 
proach the  pyramid  P  as  a  limit,  and  the  sum  a'-f-fi'-f-c'-}-  etc., 
will  approach  P  as  a  limit. 

But         a  -)-  6  -f  c  +  etc.,  =  a'  +  6'  +  c'-f-  etc.,  always. 

.-.  P=  P'.  §  275 

Q.  E.  D. 

653.  CoR.  Any  two  pyramids  having  equal  altitudes  and 
equal  bases  are  equivalent. 


POLYHEDRONS. 


53 


Proposition  XVI.     Theorem. 

654.   The  volume  of  a  triangular  pyrartiid  is  equal  to 
one  third  the  product  of  its  base  by  its  altitude. 


Given  V  the  volume,  B  the  base,  and  H  the  altitude  of  the  tri- 
angular pyramid  P-ACD. 

To  prove  F=|-BX-ff. 

Proof.  On  the  base  JB  construct  the  prism  F-ACD  with  its 
lateral  edges  II  and  =  PC. 

Pass  a  plane  through  P  and  FD. 

Then  the  prism  F~ACD  is  composed  of  three  triangular 
pyramids  P-ACD,  P-FAD,  and  P-F9^D.  c~ 

But  pyramid  P-ACD  =  pyramid  D-PF0, 

and 


§653 
Iden. 
§166 
§653 


D-PF(ff  =  P-FgD. 

Since  A  AFD  ^  A  DF^, 

.'.  P-F0D  =  P-FAD. 
.'.  P-ACD  =  P-F^D  =  P-FAD. 
.'.  the  prism  is  composed  of  three  equivalent  pyramids. 
But  the  volume  of  the  prism  =  B  x  IL  §  636 

.-.  the  volume  of  a  triangular  pyramid  =  ^  B  x  H.  fi.E.i>. 

655.   Cor.  1.     The  volume  of  any  pyramid  is  equal  to  one 
third  the  product  of  its  base  by  its  altitude. 


54 


BOOK  VII.     SOLID  GEOMETRY. 


For,  hy  drawing  diagonals  from  one  vertex  of  the  base  of  any 
pyramid  the  base  can  be  divided  into  triangles.  By  passing 
planes  through  the  vertex  of  the  pyramid  and  these  diagonals  the 
pyramid  can  be  divided  into  triangidar  pyramids. 

656.  Cor.  2.  The  volumes  of  two  pyramids  are  to  each  other 
as  the  prodxict  of  their  bases  and  altitudes  ;  pijramids  having 
equivalent  bases  and  equal  altitudes  are  equivalent. 

657.  Cor.  3.  Pyramids  having  equivalent  bases  are  to  each 
other  as  their  altitudes;  pyramids  having  equal  altitudes  are  to 
each  other  as  their  bases. 

658.  Note.  The  volume  of  any  polyhedron  may  be  determined  by 
dividing  the  polyhedron  into  pyramids,  finding  the  volume  of  each 
pyramid  and  taking  their  sum. 

Proposition  XVII.     Theorem. 

659.  A  frustum  of  a  triangular  pyramid  is  equivalent 
to  the  sum  of  three  pyraynids  ichose  common  altitude  is 
the  altitude  of  the  frustum,  and  whose  bases  are  the 
upper  base,  the  lower  base  of  the  frustum,  and  a  mean 
proportional  betiveen  them. 


Given  the  frustum  of  any  triangular  pyramid,  F-ADC,  with 
lower  base  B,  upper  base  B',  and  altitude  H. 


POLYHEDRONS  55 

To  prove,  F—ADC=  three  pyramids  whose  altitude  is  H  and 
whose  bases  are  B,  B',  and  a  mean  proportional  between  B  and  B'. 

Proof,  Divide  the  frustum,  by  passing  planes  through  E 
and  FC,  E  and  AO,  into  three  triangular  pyramids:  E-ADC, 
C-EFG,  and  E-AFG. 

Denoting  these  pyramids  by  P,  Q,  and  H,  respectively. 

(P  and  Q  have  a  common  altitude,  H.) 
Then  P  =  ^Hx  B,  and  Q^^Hx  B'.  §654 

It  remains  to  prove  that  E-AFC  or  J?  =  a  pyramid  whose 
altitude  is  H  and  whose  base  is  a  mean  proportional  between 
B  and  B'  or  V-B  x  B'. 

Because  P  and  R  have  a  common  vertex  C,  and  their  bases 
in  the  same  plane  AFED, 

.P^AAED^  §657 

R     AAEF 

But  As  AED  and  AEF  have  a  common  altitude. 

.  AAED^AD  ^g 

'  '  A  AEF     EF 

•    L  =  4R 

'  '  R     EF' 

Likewise  pyramids  Q  and  R  may  be  considered  as  having  a 
common  vertex  E,  with  their  bases  in  the  same  plane  AFOC. 

.   E^AAFC 
"  Q     A  FGC' 

AAFC^AC 
AFGC     FG' 

R^AC 
•'•  Q     FG' 


and 


56 


Because 


BOOK  Vn.     SOLID  GEOMETRY. 

A  ADC  ~  A  FGE, 

AD^AO 
EF      OF 


§  363 


.-.  B^  =  (^HxB')  X  (i  //  X  B')  =  (I  Hf  xBx  B'. 
.:E  =  ^  HVB  X  B',  or  E-AFC=  pyramid  with  altitude  H 
and  baseV5  x  B'. 


. :  frustum  F-ABC  =\  H  (B  +  B' +  VB  x  B').      q. e. d. 

660.   Formula  for  volume  of  frustum  of  a  triangular  pyra- 
mid is 

V==iH(B  +  B'+  VjB  X  B'). 


Proposition  XVIII.     Theorem. 

661.  A  frustum  of  any  pyramid  is  equivalent  to  the 
sum  of  three  pyramids  ivhose  common  altitude  is  the 
altitude  of  the  frustum,  and  lohosc  bases  are  the  upper 
base,  the  lower  base  of  the  frustum,  and  a  mean  propor- 
tional between  them. 

S  T 


POLYHEDRONS.  57 

Given  the  frustum  of  any  pyramid  Ad,  with,  its  upper  base  B', 
its  lower  base  B,  its  altitude  H,  and  its  volume  V. 


Topr(yoe  V  =  ^H{B  +  B' +  ^B  x  B'). 

Proof.     Produce  the  lateral  edges  of  the  frustum  Ad  to  meet 
inS. 

Construct  a  triangular  pyramid  T-FKG  ^v\.i\v  altitude  =  ^ 
and  base  =  ABODE  and  lying  in  the  same  plane. 

Produce  the  plane  of  abode  to  cut  the  pyramid    T-FKG 
in  fhg. 

.•.  f kg  =  abode.  §  651 

But  S-ACDE  =  T-FKG  (l),  §653 

and  S-abcde  =  T-fhg  (2). 

Subtracting  (2)  from  (1),  frustum  aD  —  frustum  fG. 


But  \o\umG  fG  =  \H{B  +  B'  +  ^BxB').  §660 


.•.V=\H{B  +  B'  +  ^BxB').  Q.E.D. 

Ex.  570.     The  diagonal  of  a  cube  is  to  its  edge  as  V3  is  to  1. 

Ex.  571.  Find  the  diagonal  of  tlie  rectangular  parallelepiped  whose 
edges  are  8  ft.,  9  ft.,  and  12  ft.,  respectively. 

Ex.  572.  Find  the  area  of  the  entire  surface  of  a  triangular  pyramid 
each  of  whose  edges  is  10  ft. 

Ex.  573.  Find  the  volume  of  a  regular  triangular  pyramid  each  of 
■whose  edges  is  8  ft. 

Ex.  574.  The  volume  of  a  truncated  parallelopiped  is  equal  to  the 
area  of  a  right  section  multiplied  by  one-fourth  the  sura  of  the  lateral 
edges. 

Ex.  575.  Find  the  volume  of  a  right  prism,  if  its  altitude  is  15  ft.,  and 
the  sides  of  its  base  are,  respectively,  10,  17,  and  21  ft. 

Ex.  576.  A  regular  pyramid,  whose  base  is  an  equilateral  triangle, 
each  side  of  which  is  12  ft.,  has  an  altitude  of  20  ft.     Find  its  volume. 


58 


BOOK  VII.    SOLID  GEOMETRY. 


Proposition  XIX.     Theorem. 

662.  A  truncated  triangular  prism  is  equivalent  to 
the  sum  of  three  pyramids  ivhose  commo7i  base  is  the 
base  of  the  prism,  and  ivhose  vertices  are  the  vertices 
of  the  inclined  section. 

D 


Given  the  truncated  triangular  prism  P,  with  base  ABC  and  in- 
clined section  D£F. 

To  prove  P  =  pyramid  F-ABC  +  pyramid  D-ABC  +  pyrcv- 
mid  E-ABC. 

Proof.  Let  the  planes  determined  by  DE  and  C,  and  by 
AC  and  E  divide  the  truncated  prism  into  three  pyramids, 
E-FDC,  E-DAC,  and  E-ABC. 

1.  E-FDC  =  B-ACF,  for  their  bases  DEC  and  AFC  are 
equal  (§  430)  and  their  altitudes  are  equal  because  their  ver- 
tices E  and  B  lie  in  the  edge  of  the  prism  which  is  II  to  the 
face  in  which  the  bases  lie. 

But  B-ACF  is  identical  with  F-ABC. 

.'.  E-FDC = F-ABC. 

2.  E-DAC  =  B-DAC,  having  same  base  and  equal  altitudes 
because  their  vertices  E  and  B  lie  in  the  edge  of  the  prism  II 
the  face  opposite. 

^nt  B-DAC  =  D-ABC,  lO^en.     .:  E-DAC  =  D-ABC. 

3.  E-ABC  has  the  required  base  and  vertex. 

.  • .  P  =  F-ABC  +  D-ABC  +  E-ABC.  q.  e.  d. 


POLYHEDRONS.  59 

663.  Cor.  1.  The  volume  of  a  right  truncated  triangular 
prism  is  equal  to  the  jyroduct  of  its  base  by  one  third  the  sum  of 
its  lateral  edges. 

664.  Cor.  2.  The  volume  of  any  truncated  triangular  prism 
is  equal  to  the  product  of  its  right  section  by  one  third  the  sum  of 
its  lateral  edges. 

Proposition  XX.     Theorem 

665.  Tetrahedrons  having  a  trihedral  angle  of  one 
equal  to  a  trihedral  angle  of  the  other  are  to  each  other 
as  the  products  of  the  edges  about  the  equal  trihedral 
angles.  f 


A  D 

Given  two  tetrahedrons  T-ABC  and  T'-DEF,  with  equal  trihedral 
i^s  at  T  and  T'  and  volumes  V  and  V'',  respectively. 

rjy                            V       TAxTBxTG 
To  prove  -— ■=—- — — . 

^  V      TDxTCxTD 

Proof.  Apply  the  tetrahedron  T-ABG  to  T-DEF  so  that 
the  equal  trihedral  ^s  T  and  T  coincide.  From  C"  and  F 
drop  ±s  upon  the  plane  TED. 

The  three  points  T,  K,  and  G  lie  in  a  straight  line,       §  578 

(the  projection  of  a  straight  line  upon  a  plane  is  a  straight  line). 
Then  with  TDE  and  TA'B'  as  the  bases  and  FG  and  CK 
the  altitudes  of  the  tetrahedrons, 

V^  ^  TA'B'  X  G'K^  TA'B'     OK  -  g^g 

V       TDExFG       TDE      FG' 


60  BOOK  VII.     SOLID   GEOMETRY. 

T>  4.  .  T'A'B'      VA'x  TB'  .  .^. 

But  -— = —. — .  §  434 

TDE       TD  X  TE 


In  the  right  similar  As  T'C'IC  and  TFG, 

CK^  TO 

FG      TF ' 


§374 


V  ^  TA  X  T'B'^  TO ^    TAxTBxTC 
'  '  V      TDxTE      TF      TBxTExTF'  q.e.d. 

666.  Polyhedrons  are  similar  if  they  have  the  same  number 
of  faces,  similar  each  to  each,  and  similarly  placed,  and  have 
their  corresponding  polyhedral  ^s  equal. 

Ex.  577.  The  altitude  of  the  fnistruni  of  a  given  pyramid  is  18  ft. 
The  lower  base  is  a  triangle  whose  sides  are,  respectively,  8  ft.,  26  ft., 
and  30  ft.  The  shortest  side  of  the  upper  base  is  4  ft.  Eind  the  volume 
of  the  finistrum. 

Ex.  578.  The  diagonal  of  one  of  the  faces  of  a  cube  is  d.  Find  the 
volume  of  the  cube. 

Ex.  579.     The  diagonal  of  a  cube  is  D.     Find  the  volume  of  the  cube. 

Ex.  580.  Find  the  lateral  area  of  a  regular  pyramid  if  it3  base  is  a 
square  16  ft.  to  the  side,  and  the  slant  height  28  ft. 

Ex.  581.  The  specific  gravity  of  mercury  being  14,  and  water  weigh- 
ing 62^  lb.  per  cubic  foot,  what  is  the  edge  of  a  cubical  box  that  would 
hold  40  lb.  of  mercury  ? 

Ex.  582.  Find  the  area  of  the  surface  of  a  regular  icosahedron,  whose 
edge  is  2  in. 

Ex.  583.  In  a  regular  pyramid  with  square  base  the  lateral  edge  is 
41  ft.  and  the  slant  height  is  40  ft.    Find  the  volume  of  the  pyramid. 


POLYHEDRONS. 


61 


PROPOSiTiojf  XXI.     Theorem. 


667,    The  homologous  edges  of  similar  polyhedrons 
are  proportional. 


Given  the  similar  polyhedrons  P  aiid  P',  with  edges  AB  and  CH 
homologous  to  A'B'  and  C'H'. 

To  prove  -— — -  =  -— — ■,. 

^  A'B'      CH' 

Proof.     Because  the  face  ABGF  ~  A'B'G'F' ; 

.    AB  ^  BG 
A'B'     B'G'' 

and  becaiase  face       BCHG  ~  B'C'HG', 

.    BG  ^  CH 
B'G'      CH' 

.    AB  ^  CH 
'  '  A'B'     CH' 


§  666 
§363 
§  666 
§363 

Q.B.D. 


Ex.  584.    The  base  of  a  regular  pyramid  is  an  equilateral  triangle 
•whose  side  is  6  ft.     The  altitude  of  the  pyramid  is  20  ft.    Find  the  volume. 


62 


BOOK  VII.     SOLID  GEOMETRY. 


668.  Cor.  1.  Any  two  homologous  lines  of  similar  polyhe- 
drons are  proportiojial  to  any  other  two  homologous  lines  of  the 
polyhedrons. 

669.  Cor.  2.  Any  two  homologous  faces  of  similar  polyhe- 
drons are  proportional  to  the  squares  of  any  two  homologous  lines 
of  the  polyhedrons,  and  the  total  surface  of  two  similar  polyhedrons 
are  proportional  to  the  squares  of  any  two  homologous  lines  of  the 
polyhedrons. 

Proposition  XXII.     Theorem. 

670.  Similar  tetrahedrons  are  to  each  other  as  the 
cubes  of  their  homologous  edges. 


B  B' 

Given  two  similar  tetrahedrons  T  and  T',  whose  volumes  are  V 
and  V,  and  whose  altitudes  are  H  and  H',  respectively. 

V      A& 


To  prove 

r     A'B" 

Proof. 

V       base  ABC  X 
V'     bsiseA'B'C'x 

H 
W 

§  656 

But 

base  ABC       AB^ 
base  A'B'C     a'B'^ 

§  669 

And 

H      AB 

H'     A'B' 

§  668 

.    V  _   AB'xAB 
V     A'B''  X  A'B' 

Alf 
A'B'' 

Q.  E.  D. 

POLYHEDRONS. 
Proposition  XXIII.     Theorem. 


63 


671.  Similar  polyhedrons  may  he  divided  into  the 
same  number  of  tetrahedrons  similar  each  to  each  and 
similarly  placed. 


B'  C  B  G 

Given  the  two  similar  polyhedrons,  P  and  P'. 

To  prove  P  and  P'  may  be  divided  into  the  same  number  of 
tetrahedrons  similar  each  to  each,  and  similarly  placed. 

Proof.  Take  any  homologous  trihedral  ^s  in  P  and  P',  as  B 
and  B',  and  through  points  G,  A,  C  pass  a  plane ;  also  through 
points  G',  A',  C  pass  a  plane.  Then  in  the  two  tetrahedrons 
thus  cut  off,  G-ABC  and  G'-A'B'C, 

AG       BG       CG       AC 


A'G'     B'G'     CG'     A'C' 


§  363 


.'.  the  faces  BAG,  BAC,  and  BGC  are  similar  to  B'A'G', 
B'A'C,  and  B'G'C  respectively.  §  376 

.-.  the  homologous  faces  of  these  tetrahedrons  are  similar. 

§  3G9 

But  the  homologous  trihedral  ^s  of  these  tetrahedrons  are 
equal.  §  595 

.-.  tetrahedron  (^-^JSC~  tetrahedron  G'-A'B'C.  §  666 

After  removing  tetrahedron  G-ABC  from  P  and  G'-A'B'C 
from  P'  the  polyhedrons  which  remain  will  be  similar,  for 
their  faces  are  similar  and  the  polyhedral  ^s  are  equal.     By 


64  BOOK  VII.     SOLID  GEOMETRY. 

this  process  P  and  P'  may  be  divided  into  the  same  number  of 
tetrahedrons,  similar  each  to  each  and  similarly  placed,    q.e  d. 

672.  Cor.  Tlie  volumes  of  any  two  similar  polyhedrons  are 
to  each  other  as  the  cubes  of  any  two  homologous  lines  of  the 
polyJiedrons. 

673.  A  regular  polyhedron  has  all  its  faces  congruent  regular 
polygons  and  all  its  polyhedral  angles  congruent. 

Pboposition  XXIV.     Theorem. 

674.  Only  five  regular  polyhedrons  are  possible. 
Given,  congruent  regular  polygons  of  any  number  of  sides. 

To  prove,  only  five  regular  polyhedrons,  with  congrxient  regular 
polygons  for  faces,  can  he  constructed. 

Proof.  A  polyhedron  '^  must  have  at  least  three  faces  and 
the  sum  of  the  face  ^s  must  be  <  360°. 

1.  With  equilateral  As  where  each  ^  is  60°; 

60°  X  3  =  180°;  60°  X  4=240°;  60°  x  5=300°;  but  60°  x  6=360°. 

.-.  only  three  regular  polyhedrons  can  be  formed  with  equi- 
lateral As  for  faces. 

2.  With  squares  where  each  ^  is  90°, 

90°  X  3  =  270° ;  but  90°  x  4  =  360°. 

.'.  only  one  regular  polyhedron  can  be  formed  with  squares 
for  faces. 

3.  With  regular  pentagons  where  each  ^  is  108°. 

108x3  =  324°;  but  108°  X  4  =  432° 

.'.  only  one  regular  polyhedron  can  be  formed  with  regular 
pentagons  for  faces. 

4.  But  with  regular  hexagons  where  each  ^  is  120°,  because 
120°  X  3  =  360°,  no  regular  polyhedron  can  be  formed.     Hence 


POLYHEDRONS. 


65 


no  regular  polyhedron  can  be  formed  with  polygons  having 
more  than  five  sides. 

.-.  only  five  regular  polyhedrons  are  possible.  q.e.  d. 

675.  Note.  The  five  regular  polyhedrons  are  the  tetrahedron,  the 
octahedron,  and  the  icosahedron  from  equilateral  triangles;  the  hexahe- 
dron or  cube  from  squares  ^and  the  dodecahedron  from  pentagons. 

676.  These  regular  polyhedrons  may  be  formed  by  cutting 
out  cardboard  as  indicated  in  the  following  diagrams.  Cut 
entirely  through  on  the  full  lines,  half  through  on  the  broken 
lines,  and  bring  the  edges  together.  The  edges  may  be  held 
in  place  by  pasting  narrow  strips  of  paper  over  them. 


66 


BOOK  Vn.     SOLID   GEOMETRY. 


CYLINDERS. 

677.  A  cylindrical  surface  is  a  curved  surface  generated  by  a 
moving  straight  line  that  constantly  touches  a  given  curve  and 
is  always  parallel  to  a  fixed  straight 

line. 

The  moving  line  is  called  the  gen- 
eratrix and  the  given  curve  the  direc- 
trix. The  generatrix  in  any  position 
is  called  an  element  of  the  cylindrical 
surface. 

678.  A  cylinder  is  a  solid  bounded 
by  a  cylindrical  surface,  called  the 
lateral  surface  and  two  parallel  planes 
which  are  the  bases  of  the  cylinder. 

679.  The  altitude  of  a  cylinder  is 
the  perpendicular  distance  between 
the  bases. 

680.  Because  parallel  lines  included 

between  parallel  planes  are  all  equal,  the  elements  of  a  cylin- 
drical surface  are  all  equal. 

681.  A  circular  cylinder  is  a  cylinder  whose  bases  are  circles. 
The   term    "cylinder,"   as   hereafter  used, 
will  mean  circular  cylinder,  as  the  circle  is 
the  only  curve  discussed  in  elementary  plane 
geometry. 

682.  A  right  cylinder  is  a  cylinder  whose 
elements  are  perpendicular  to  its  bases. 
An  oblique  cylinder  is  one  whose  elements 
are  oblique  to  its  bases. 

A  right  cylinder  is  called  a  cylinder  of 
revolution  because  it  may  be  generated  by  the  revolution  of  a 


CYLINDERS. 


67 


rectangle  about  one  of  its  sides  as  an  axis.     The  radius  of  the 

base    is    the    radius   of   the 

cylinder. 

683.  Similar  cylinders    of 

revolution  are  cylinders  gen- 
erated by  similar  rectangles 
revolving  about  correspond- 
ing sides  as  axes. 

684.  A  plane  is  tangent  to 

a  cylinder   when    it   passes 

through   one  element  of  the   cylinder   but   does   not  cut  it. 

685.  A  prism  is  inscribed  in  a  cylinder  when  its  base  is  a 
polygon  inscribed  in  the  base  of  the  cylinder  and  its  lateral 
edges  are  elements  of  the  cylinder. 

686.  A  prism  is  circumscribed  about  a  cylinder  when  its  bases 
are  polygons  circumscribed  about  the  bases  of  the  cylinder 
and  its  lateral  faces  are  tangent  to  the  cylinder. 

687.  A  right  section  of  a  cylinder  is  the  figure  formed  by 
the  intersection  of  a  plane  with  the  cylinder  perpendicular  to 
its  elements.     The  right  section  of  a  cylinder  is  a  circle. 

688.  Because  the  circle  is  the  limit  of  the  perimetersjof-regu- 
Ja£_jnscribed  and  circumscribed   polygons  and  the  area  of  a- 

circle   is^  the  limit  of   the  areas  of  these  polygons  when  the 
iminBer  of  their  sides  is  indefinitely  increased,  §  488,  hence: 

1.  The  circle  of  a  right  section  of  a  cylinder  is  the  limit  of 
the  perimeter  of  a  right  section  of  an  inscribed  or  circum- 
scribed prism. 

2.  The  lateral  area  of  a  cylinder  is  the  limit  of  the  lateral 
area  of  the  inscribed  or  circumscribed  prism. 

3.  The  volume  of  a  cylinder  is  the  limit  of  the  volume  of 
an  inscribed  or  circumscribed  prism. 


68  BOOK  VII.     SOLID   GEOMETRY. 

Proposition  XXV.     Theorem. 
689.     Tlie  lateral  area  of  a  cylinder  is  equal  to  the 
prodiLct  of  the  circle  of  a  right  section  by  an  element. 


Given  the  cylinder  AB,  with  lateral  area  S,  circle  of  a  right  section 
C,  and  element  H. 

To  prove  S=CxH. 

Proof.  Inscribe  in  the  cylinder  a  regular  prism,  with  lateral 
area  S',  and  perimeter  of  a  right  section  P.  Its  lateral  edge 
is  E. 

Then  S'  =  PxII.  §  616 

Let  the  number  of  lateral  faces  of  the  prism  be  indefinitely 
increased. 

Then  S'  approaches  /S'  as  a  limit.  §  688,  2 

P  approaches  C  as  a  limit  §  688,  1 

and  Px  H approaches  C  X  H a.s  a.  limit.  §  277 

But  S'  =  PxH  always.  §  616 

.-.  S=CxH.  §275 

Q.  E.  D. 

690.  Cor.  1.  The  lateral  area  of  a  cylinder  of  revolution  is 
equal  to  the  circle  of  the  base  multiplied  by  the  altitude. 

691.  Cor.  2.  If  8  is  the  lateral  area,  T  the  total  area,  and 
H  an  element,  then 

S  =  2irBxH. 

T=2-nBx  H+  2  irR^  =2  irR{H-\-  B). 


CYLINDERS. 


69 


692.    Cor.  3.     Lateral  areas  or  total  areas  of  two  similar 
cylinders  are  to  each  other  as  the  squares  of  their  like  dimensions. 


Pbopositiox  XXVI.    Theorem. 

693.    The  volume  of  a  cylinder  is  equal  to  the  product 
of  its  base  and  altitude. 


Given  the  cylinder  AB,  with  volume  V,  base  B,  and  altitude  H. 

To  prove  V=B  x  H. 

Proof.     Inscribe  in  the  cylinder  a  regular  prism  with  volume 
V  and  base  B'.     Its  altitude  is  equal  to  II. 

Then  V'=B'xH.  §636 

Let  the  number  of  the  lateral  faces  of  the  prism  be  indefi- 
nitely increased. 

Then  "P  approaches  F  as  a  limit, 

B'  approaches  jB  as  a  limit, 
and  B'  X  H  approaches  J5  X  ^  as  a  limit. 

.    But  F  =  5' X  ^  always. 

.-.  V=BxH. 


§  688,  3 

§  688,  1 

§  277 

§  636 

§275 

Q.E  D. 


694.  Cor.  1.     If  the  radius  of  the  cylinder  is  R,  then 

V=  ttR'H. 

695.  Cor.  2.     Volumes  of  two  similar  cylinders  are  to  each 
other  as  the  cubes  of  their  like  dimensions. 


70 


BOOK  VIL     SOLID  GEOMETRY. 


CONES. 

696.  A  conical  surface  is  a  curved  surface 
generated  by  a  Inoving  straight  line  that 
constantly  touches  a  given  curve  and  passes 
through  a  fixed  point.  The  moving  line  is 
called  the  generatrix,  the  fixed  point,  the 
vertex,  and  the  given  curve,  the  directrix. 

The  generatrix  in  any  position  is  called 
an  element  of  the  conical  surface. 

697.  The  conical  surface  may  consist  of 
two  parts,  one  above  and  the  other  below 
the  vertex,  called  the  upper  and  lower  nappes, 
respectively. 

698.  A  cone  is  a  solid  bounded  by  a  coni- 
cal surface,  called  the  lateral  surface  and  a 
plane  which  is  the  base  of  the  cone. 

699.  The  altitude  of  a  cone  is  the  per- 
pendicular from  the  vertex  to  the  plane  of 
the  base. 

700.  A  circular  cone  is  a  cone  whose  base 
is  circular.     The  term  "  cone  "  as  hereafter 
used,  will  mean  a  circular 
cone  of  one  nappe. 

701.  The  axis  of  a  cone 
is  the  straight  line  from 
the  vertex  to  the  center  of 
the  base, 

702.  A  right  cone  is  a 
cone  in  which  the  axis  is 
perpendicular  to  the  base.  A  right  cone  is  also  called  a  cone 
of  revolution  because  it  may  be  generated  by  revolving  a  right 
triangle  about  one  of  its  legs  as  an  axis. 


CONES.  71 

703.  The  slant  height  is  an  element  of  the  conical  surface. 

704.  Similar  cones  of  revolution  are  cones  generated  by  simi- 
lar right  triangles  revolving  about  corresponding  legs. 

705.  A  plane  is  tangent  to  a  cone  when  it 
touches  it  in  one  element  of  the  cone  but 
does  not  cut  it. 

706.  A  pyramid  is  inscribed  in  a  cone 
when  its  base  is  a  polygon  inscribed  in  the 
base  of  the  cone  and  its  lateral  edges  are 
elements  of  the  cone. 

707.  A  pyramid  is  circumscribed  about 
a  cone  when  its  base  is  a  polygon  circum- 
scribed about  the  base  of  the  cone,  and  its 
lateral  faces  are  tangent  to  the  cone. 

708.  When  the  number  of  sides  of  regu- 
lar inscribed  or  circumscribed  polygons  is 
indefinitely  increased,  §§  488,  688, 

1.  The  circle  of  a  right  section  of  a  cone  is  the  limit  of  the 
perimeter  of  a  right  section  of  an  inscribed  or  circumscribed 
pyramid. 

2.  The  lateral  area  of  a  cone  is  the  limit  of  the  lateral  area 
of  an  inscribed  or  circumscribed  pyramid. 

3.  The  volume  of  a  cone  is  the  limit  of  the  volume  of  an 
inscribed  or  circumscribed  pyramid. 

709.  The  axis  of  a  right  cone  is  its  altitude. 

710.  The  elements  of  a  right  cone  are  all  equal. 

711.  The  frustum  of  a  cone  is  the  part 
of  a  cone  contained  between  its  base  and 
a  plane  parallel  to  its  base.  The  base  of 
the  cone  is  the  lower  base  of  the  frustum 
and  the  section  made  by  the  plane  parallel 
to  the  base  is  the  upper  base  of  the  frustum. 


72 


BOOK  VII.     SOLID  GEOMETRY. 


Proposition  XXVII.     Theorem. 

712.  The  lateral  area  of  a  cone  of  revolution  is  equal 
to  half  the  product  of  the  slant  height  hy  the  circle  of 
the  base. 


Given  V-EFG  a  cone  of  revolution,  L  its  slant  height,  C  the  circle 
of  its  base,  and  S  its  lateral  area. 


To  prove 


S  =  \-LxC. 


Proof.  Circumscribe  about  the  cone  a  regular  pyramid,  de- 
noting its  lateral  area  by  S'  and  the  perimeter  of  its  base 
by  P. 

Then  S'  =  ^LxP.  §  646 

Let  the  number  of  lateral  faces  of  the  pyramid  be  indefi- 
nitely increased. 

Then  S'  will  approach  /S  as  a  limit,  §  708,  2 

P'  will  approach  P  as  a  limit,  §  708,  1 

^nd  \  Lx  P  will  approach  \L  x  C  as  a  limit.  §  277 

But  S'  =  ^LxP  always. 

.'.  S  =  \L  X  C.  Q.E.D. 


CONES.  73 

713.  Cor.  1.  If  R  is  the  radius  of  a  cone  and  T  is  its  total 
surface,  because  the  circle  of  the  base  is  2  tH,  and  the  area  is 
irR\ 

S=ix2TrRxL  =  ttRL.     T=  tRL  +  7ri22  =  TrR{L  +  R). 

714.  Cor.  2.  The  lateral  areas  or  the  total  areas  of  two 
similar  cones  are  to  each  other  as  the  squares  of  their  like  dimen- 
sions. 


Proposition  XXVIIL 

715.  T7ie  lateral  area  of  the  frustum  of  a  right  cone 
is  equal  to  half  the  product  of  the  slant  height  of  the 
frustum  hy  the  sum.  of  the  circles  of  the  bases. 

Given  L'  the  slant  height,  c  the  upper  base 
of  the  frustum,  C  the  lower  base,  and  S'  the 
lateral  area. 

To  prove  S'  =  ^  (c  +  C)L'. 

Proof.  The  lateral  area  of  the  frustum 
of  the  inscribed  pyramid  is 

S  =  i{p  +  P)L.  §647 

"When  the  number  of  lateral  faces  of  the  inscribed  frustum 
of  the  pyramid  is  indefinitely  increased,  F  api)roaches  C,  p  ap- 
proaches c,  L  approaches  L',  and  aS  approaches  S'  as  a  limit. 

.:  S'  =  ^{c+C)L'.  Q.E.D. 

716.  Cor.  Tlie  lateral  area  of  the  frustum  of  a  cone  of  revo- 
lution is  equal  to  the  j)roduct  of  the  circle  of  a  section  midway 
between  the  bases,  by  the  slant  height. 


74  BOOK  VII.    SOLID  GEOMETRY. 

Proposition  XXIX.     Theorem. 

717.    The  volume  of  a  cone  is  equal  to  one  third  the 
product  of  its  hose  hy  its  altitude. 


Given  V  the  volume,  B  the  base,  and  H  the  altitude  of  the  cone 
V-EFG. 

To  prove  V=\BxH. 

Proof.  Inscribe  in  the  cone  a  regnlar  pyramid,  denoting  its 
volume  by  V,  its  base  by  B',  and  //  will  be  its  altitude. 

Then  V  =  \B' x  H.  §655 

Let  the  number  of  lateral  faces  of  the  pyramid  be  indefi- 
nitely increased. 

Then  V  approaches  F  as  a  limit.  §  708,  3 

Also  Bf  approaches  B  as  a  limit.  §  708,  1 

But  V  =  \B'  xH  always.  §  656 

.:V=\BxH.  §275 

Q.E.D. 

718.  CoK.  1.     If  ttH^  =  area  of  the  base, 

719.  Cor.  2.  The  volumes  of  similar  cones  are  to  each  other 
as  the  cubes  of  their  like  dimensions. 


CONES.  75 

Proposition  XXX,     Theorem. 

720.  A  frustum  of  a  cone  is  equivalent  to  the  sum  of 
three  cones  of  the  altitude  of  the  frustum,  and  ivhose 
bases  are  the  upper  base,  the  lower  base  of  the  frustum, 
and  a  mean  proportional  between  them. 


Given  V  the  volume,  b  the  upper  base,  B  the  lower  base,  and  H  the 
altitude  of  the  frustum  F. 


To  prove  V=  \  H{h  +B+  y/b  x  B). 

Proof.  Inscribe  in  F  the  frustum  of  a  regular  pyramid, 
denoting  its  volume  by  V,  its  bases  by  b'  and  B'.  II  will  be 
its  altitude. 

Then  F  =  ^  7/(6 '  +  5'  +  V&'  X  B').  §  661 

Let  the  number  of  lateral  faces  of  the  inscribed  frustum  be 
indefinitely  increased. 

Then  V  will  approach  Fas  a  limit.  §  708,  3 

Also  B'  will  approach  jB  as  a  limit.  §  708,  1 

And  6'  will  approach  6  as  a  limit.  §  708,  1 

But  V  =i  H{h'  +B'  +  Vb'  X  B')  always.      §  661 

.-.  F  =  i  H(b  +  5  +  ^b^B).  §  275 

Q.  E.  D. 

721.  If  r  and  B  denote  the  radii  of  the  upper  and  lower 
bases  respectively,  then  itr^  and  wR^  denote  their  areas,  and 


76  BOOK   VII.     SOLID   GEOMETRY 

Ex.  585.  Find  the  volume  of  a  cube  whose  entire  surface  is  54  sq.  ft. 

Ex.  586.  The  radius  of  the  lower  base  of  a  frustum  of  a  cone  is  21,  the 
radius  of  the  upper  base  10,  and  the  slant  height  (Jl.     Find  its  volume. 

Ex.  587.  A  chimney  60  ft.  high  is  in  the  shape  of  the  frustrum  of  a 
cone.  Its  lower  diameter  is  28  ft.  and  the  upper  diameter  20  ft.  The 
conical  flue  has  the  lower  diameter  12  ft.  and  the  upper  diameter  6  ft. 
Find  the  volume  of  the  chimney. 

Ex.  588.  The  volumes  of  two  similar  prisms  are  to  each  other  as  5  to  6. 

What  is  the  ratio  of  their  surfaces  ? 

Ex.  589.  Find  the  volume  of  a  frustum  of  a  regular  quadrangular 
pyramid,  the  sides  of  whose  bases  are  10  and  0  and  whose  altitude  is  12. 

Ex.  590.  The  radius  of  the  lower  base  of  the  frustum  of  a  cone  is  34, 
the  radius  of  the  upper  base  20,  and  the  altitude  48.     Find  its  lateral  area. 

Ex.  591.  The  altitude  of  the  Great  Pyramid  is  488  ft.  and  its  base  is 
764  ft.  square.     What  is  its  volume  ? 

Ex.  592.  The  altitude  of  a  pyramid  is  9  ft.  and  its  base  is  a  rhombus 
whose  diagonals  are,  resijectively,  10  and  12  ft.     What  is  its  volume  ? 

Ex.  593.  The  lateral  edge  of  a  pyramid  is  10  ft.  and  its  inclination  to 
the  base  is  30°.  The  base  is  an  equilateral  triangle  whose  side  is  12  ft. 
Find  the  volume  of  the  pyramid. 

Ex.  594.  Find  the  volume  of  a  truncated  right  triangular  prism,  the 
sides  of  whose  base  are,  respectively,  13,  14,  and  15  ft.,  and  whose  lateral 
edges  are  6,  8,  and  10  ft.,  respectively. 

Ex.  595.  Find  the  volume  of  the  cube,  in  which  the  diagonal  of  each 
face  is  16  in. 

Ex.  596.  The  base  of  a  right  pyramid  is  a  regular  hexagon  of  side  18 
in.  and  the  lateral  faces  are  inclined  to  the  base  at  an  angle  of  60"^. 
Find  the  volume. 


Book  VIII. 


THE   SPHERE. 


722.  A  sphere  is  a  solid  bounded  by  a  surface,  all  points  of 
which  are  equidistant  from  a  point  within,  called  the  center. 

723.  The  radius  of  a  sphere  is  a  straight  line  from  the  center 
to  any  point  on  the  surface. 

724.  The  diameter  of  a  sphere  is  a  straight  line  through  the 
center  terminated  at  both  ends  by  the  surface. 

725.  It  follows  from  the  definition  of  the  sphere  that  all 
radii  of  the  same  sphere  or  of  equal  spheres  are  equal,  that  all 
diameters  of  the  same  sphere  or  of  equal  spheres  are  equal,  and 
that  spheres  are  equal  if  their  radii  or  their  diameters  are  equal. 

726.  A  sphere  may  be  gener- 
ated by  the  revolution  of  a  semi- 
circle about  its  diameter  as  an 
axis. 

727.  A  line  or  a  plane  which 
has  one,  and  only  one,  point  in 
common  with  the  surface  of  a 
sphere  is  tangent  to  the  sphere. 

The  sphere  is  then,  also,  tangent  to  the  line  or  the  plane. 

728.  Two  spheres  whose  surfaces  have  one,  and  only  one, 
point  in  common  are  tangent  to  each  other. 

729.  A  polyhedron  is  inscribed  in  a  sphere  when  all  its  ver- 
tices are  in  the  surface  of  the  sphere.  The  sphere  is  then  cir- 
cumscnbed  about  the  polyhedron. 

77 


78  BOOK  Vin.     SOLID  GEOMETRY. 

730.  A  polyhedron  is  circumscribed  about  a  sphere  when  all 
its  faces  are  tangent  to  the  sphere.  The  sphere  is  then  ii\r 
scribed  in  the  polyhedron. 

Proposition  I.     Theorem. 

731.  The  intersection  of  a  plane  and  the  surface  of  a 
sphere  is  a  circle. 


Given  ABC,  a  section  made  by  a  plane  cutting  the  sphere  whose 
center  is  0. 

To  prove  ABC  is  a  O. 

Proof.  Let  OQ  be  ±  plane  ABC. 

From  A  and  B,  any  two  points  in  the  boundary  of  the  sec- 
tion ABC,  draw  AO  and  BO,  and  draw  also  AQ  and  BQ. 
Then  in  rt.  As  OAQ  and  BOQ, 

OQ  =  OQ.  Iden. 

AO  =  BO.  §725 

..AQ  =  BQ.  §§122,85 

But  A  and  B  are  any  two  points  on  the  boundary  of  the 

section  ABC  .  ^^^  ^^  ^  q  §  208 

Q.  E.  D. 

732.  CoR.  1.  The  line  from  the  center  of  a  sphere  to  the  cen- 
ter of  a  circle  of  the  sphere  is  perpendicular  to  the  plane  of  tJie 
circle. 

733.  Def.  a  great  circle  of  a  sphere  is  a  section  of  the 
sphere  made  by  a  plane  that  passes  through  the  center  of  the 
sphere. 


THE   SPHERE.  79 

734.  Def.  a  small  circle  of  a  sphere  is  a  section  of  the 
sphere  made  by  a  plane  that  does  not  pass  through  the  center 
of  the  sphere. 

735.  Def.  The  diameter  of  a  sphere  perpendicular  to  the 
plane  of  a  circle  of  the  sphere  is  called  the  axis  of  the  circle, 
and  the  extremities  of  the  diameter  are  called  the  poles  of  the 
circle. 

736.  CoR.  2.  A  great  circle  has  the  same  center  and  the  same 
radius  as  the  sphere,  hence  all  great  circles  of  the  same  sphere  or  of 
all  equal  spheres  are  equal. 

737.  CoR.  3.  A  great  circle  bisects  the  sphere  and  the  surface 
of  the  sphere, 

For,  if  the  two  parts  into  which  it  divides  the  sphere  be  so 
placed  that  their  plane  surfaces  coincide,  then  the  curved  sur- 
faces must  coincide,  otherwise  there  would  be  points  in  the 
surface  of  the  sphere  at  different  distances  from  the  center. 

738.  Def.  The  distance  between  any  two  points  on  the 
surface  of  a  sphere  is  the  arc  of  a  great  circle,  not  greater  than 
a  semicircle,  that  joins  the  points. 


Ex.  697.  Of  two  given  cubes  the  diagonal  of  the  first  is  three  times 
that  of  the  second.     What  is  the  ratio  of  their  volumes  ? 

Ex.  598.  The  edges  of  a  given  rectangular  parallelepiped  are,  respec- 
tively, 9  ft.,  24  ft.,  and  32  ft.  What  is  the  volume  of  a  similar  parallelo- 
piped,  whose  diagonal  is  82  ft.  ? 

Ex.  599.  A  pyramid  has  an  altitude  of  26  ft.  At  what  distance  from 
the  base  must  it  be  cut  by  a  plane  parallel  to  the  base  that  the  frustum 
may  be  half  the  pyramid  ? 

Ex.  600.  If  the  area  of  the  entire  surface  of  a  tetrahedron  is  200  sq.  ft. 
and  the  altitude  60  ft,,  what  is  the  altitude  of  a  similar  tetrahedron 
whose  entire  surface  is  800  sq.  ft.  ? 


80  BOOK  VIII.     SOLID  GEOMETRY. 

Proposition  II.     Theorem. 

739.  All  points  on  a  circle  of  a  sphere  are  equidis- 
tant frmn  either  of  its  poles. 


Given  two  points,  D  and  £  on  the  O  D£F,  and  A  and  B,  the 
poles  of  the  O  DEF. 

To  prove  the  great  circle  arcs  AD  and  AE  are  equal,  and  the 
great  circle  arcs  BD  and  BE  are  equal. 

Proof.     The  straight  lines  AD  and  AE  are  equal.  §  533 

.-.  arc  AD  =  arc  AE.  §  234 

Similarly,  arc  BD  =  arc  BE.  Q.  b.  d. 

740.  Def.  The  distance  on  the  surface  of  a  sphere  from 
the  nearer  pole  of  a  circle  to  any  point  of  the  circle  is  called 
the  polar  distance  of  the  circle. 

741.  Cor.  Hie  polar  distance  of  a  great  circle  of  a  sphere  is 
the  qvxidrant  of  a  great  circle. 

Ex.  601.  The  base  of  a  regular  pyramid  is  an  equilateral  triangle 
whose  side  is  18  ft.  The  slant  height  of  the  pyramid  is  30  ft.  Find  the 
volume. 

Ex.  602.  Of  two  similar  pyramids  the  entire  surface  of  the  first  is 
four  times  that  of  the  .^second.  What  is  the  ratio  of  the  volume  of  the 
first  to  that  of  the  second  ? 


THE  SPHERE. 


81 


Proposition  III.     Problem. 
742.     To  construct  the  radius  of  a  material  sphere. 


Let  MNP  represent  a  material  sphere. 
Required  to  construct  its  radius. 

Construction.  Take  any  two  points,  A  and  B,  on  the  surface 
of  the  sphere,  as  poles,  and  with  the  same  radius  construct  two 
arcs  intersecting  each  other  at  C;  then  with  the  same  poles  and 
with  other  equal  radii,  construct  two  arcs  intersecting  at  D ; 
finally,  with  still  the  same  poles  and  with  other  equal  radii, 
construct  two  arcs  intersecting  at  E. 

The  three  points,  C,  D,  and  E,  thus  determined,  determine 
the  plane  that  is  the  perpendicular  bisector  of  the  straight 
line  joining  A  and  B.  §§  523,  538 

.".  the  plane  GDE  passes  through  the  center  of  the  sphere. 

§§  722,  538 

.♦.  C,  D,  E  lie  on  the  great  circle  of  the  sphere.  §  733 

Construct  a  plane  triangle  OD'E',  whose  sides  are  equal, 
respectively  to  CD,  DE,  and  CE.  §  303 

Circumscribe  the  circle  0  about  the  plane  triangle  CD'E'. 

§306 

Then  Q  0  =  great  O  of  the  sphere.  §  246 

.'.  OC,  the  radius  of  O  0  =  radius  of  great  O  (§  246)  =  ra- 
dius of  the  sphere,  §  213 

Q.B.F. 


82 


BOOK  Vm.     SOLID   GEOMETRY 


Propositiox  IV.     Theorem. 

743.    A  plane  perpendicular  to  a  radius  at  its  outer 
extremity  is  tangent  to  the  sphere. 


Given  0,  the  center  of  a  sphere,  MN,  a  plane  A.  radius  OA  at  A. 
To  prove  the  plane  MN  is  tangent  to  the  sphere. 

Proof.     Let  B  be  any  other  point  except  A  in  the  plane  MN. 
Then  OB  >  OE.  §  531 

.'.  the  point  B  is  without  the  sphere.  §  722 

But  B  is  any  point  in  the  plane  MN  other  than  A. 
.'.  the  plane  MN  is  tangent  to  the  sphere. 

Q.  E.  D. 

744.  Cob.  1.     A  plane  tangent  to  a  sphere  is  perpendicular  to 
the  radius  drawn  to  the  point  of  tangency. 

745.  Cor.  2.     A  line  in  a  tangent  plane  drawn  through  the 
point  of  tangency  is  tangent  to  the  sphere  at  that  point. 

746.  Cor.  3.     A  line  tangent  to  a  circle  of  a  sphere  lies  in  the 
plane  that  is  tangent  to  the  sphere  at  the  point  of  contact. 

Ex.  603.    If  the  volumes  of  two  similar  prisms  are  to  each  other  as 
8  to  27,  what  is  the  ratio  of  their  altitudes  ? 


THE  SPHERE.  83 

Proposition  V.     Theorem. 

747.  Through  any  four  points,  not  all  in  the  same 
plane,  one,  and  only  one,  sphere  may  he  passed. 

A 


c 

Given  A,  B,  C,  and  D,  four  points,  not  all  in  the  same  plane. 

To  prove  one  sphere,  and  only  one,  may  he  passed  through  A, 
B,  C,  and  D. 

Proof.  Let  F  and  G  be  the  centers  of  Os  circumscribing 
As  BCD  and  ACD,  respectively.  Let  FK  be  ±  tlie  plane  BCD 
and  GH  ±  the  plane  ACD. 

Then  every  point  in  FK  is  equidistant  from  B,  C,  and  D, 
and  every  point  in  GH  is  equidistant  from  A,  C,  and  Z).  §  533 

Join  F  and  G  to  E,  the  mid-point  of  CD. 

Then  FE  and  GE  are  each  ±  CD.  §  240 

.-.  the  plane  GEF 1.  CD.  §  535 

.-.  the  plane  GEF  A.  planes  BCD  and  ACD.        §  567 

Then,  since  GH  is  X  plane  ACD  by  constniction,  GH  lies  in 
the  plane  GEF.  §  569 

Similarly,  FK  lies  in  the  plane  GEF.  Therefore  ±s  GH  and 
and  FK  lie  in  the  same  plane  and  being  ±  to  non-parallel 
planes,  they  meet  in  some  point,  as  0. 

.•.  0  lies  in  ±  GH  and  FK  equidistant  from  B,  C,  and  D, 
and  from  A,  C,  and  D.  §  533 

.'.  0  is  equidistant  from  A,  B,  C,  and  D. 


84  BOOK   VIII.     SOLID   GEOMETRY. 

Hence  the  sphere  whose  center  is  0  and  radius  OA  will  pass 
through  A,  B,  C,  and  Z>. 

Again,  since  the  center  of  any  sphere  through  A,  B,  C,  and 
D  must  lie  in  GH  and  FK  (§  534),  their  intersection  0  is  the 
center  of  the  only  sphere  that  will  pass  through  the  four  points 
A,  B,  C,  and  D.  q.e.d. 

748.  CoR.  Four  points  not  all  in  the  same  plane  determine  a 
sphere. 

SPHERICAL  ANGLES. 

749.  Def.  The  angle  formed  by  two  intersecting  curves  is 
the  angle  formed  by  the  tangents  to  the  curves  at  the  point  of 
intersection. 

750.  Def.  The  angle  formed  by  two  intersecting  great 
circles  of  a  sphere  is  called  a  spherical  angle. 

Proposition  VI.     Theorem. 

751.  A  spherical  angle  is  'measured  hy  the  are  of  a 
great  circle  described  from  the  vertex  as  a  pole  and  in- 
cluded between  its  sides,  or  its  sides  produced. 


Given  the  great  Os  BCA  and  BDA,  intersecting  at  A,  and  CD,  the 
arc  of  a  great  O  described  with  A  as  a  pole. 

To  prove  the  spherical  angle  CAD  is  measured  by  tJie  arc  CD. 


THE   SPHERE.  85 

Proof.     Draw  radii  OC  and  OD  and  tangents  AE  and  AF. 

Arcs  AC  and  AD  are  quadrants.  §  216 

OG  and  OD  each  ±  0^,  and  ^  COD  ^  ^  EAF.     §  555 

.'.  the  spherical  ^  CAD  is  measured  by  the  ai'c  CD.  q.e.  d. 

^52.   Cor.    A  spherical  angle  has  the  same  measure  as  the 
dihedral  angle  formed  by  the  planes  of  the  two  circles. 

SPHERICAL  POLYGONS. 

753.  Def.  a  spherical  polygon  is  a  portion  of  the  surface  of 
a  sphere  bounded  by  three  or  more  great  circles. 

The  bounding  arcs  are  called  the  sides,  their  points  of  inter- 
section, the  vertices,  and  the  spherical  angles  formed  by  the 
slides,  the  angles  of  the  spherical  polygon. 

754.  Def.  The  diagonal  of  a  spherical  polygon  is  the  arc 
of  a  great  circle  drawn  between  two  non-consecutive  vertices. 

755.  The  planes  of  the  sides  of  a  spherical  polygon  form 
a  polyhedral  angle  at  the  center  of  the  sphere.  A  spherical 
polygon  is  .convex  if  the  corresponding  polyhedral  angle  is  con- 
vex. Unless  stated  otherwise  a  spherical  polygon  is  assiimed 
to  be  convex.  From  any  property  of  polyhedral  angles  may  be 
inferred  an  analogous  property  of  spherical  polygons,  and  con- 
versely. 

756.  The  measures  of  the  sides  of  a  spherical  polygon  are 
usually  expressed  in  degrees. 

757.  Def.  Two  spherical  polygons  are  vertical  when  their 
corresponding  polyhedral  angles  are  vertical. 

758.  Two  spherical  polygons  are  symmetrical  when  the 
parts  of  one  are,  respectively,  equal  to  the  parts  of  the  other 
and  arranged  in  reverse  order. 


86  BOOK  Vin.    SOLID  GEOMETRY. 

Proposition  VII.    Theorem. 
759.    Two  vertical  spherical  triangles  are  symmetrical. 


Given  two  vertical  spherical  triangles,  ABC  and  A'B'C . 

To  prove  sphencal  As  ABC  and  A'B'C  are  symmetrical. 

Proof.     Let  0  be  tlie  center  of  the  sphere. 

Plane  As  AOC  and  A'OC  have  two  sides  and  the  included 
^s  respectively  equal,  and  are,  therefore,  congruent.    §§  65,  91 

.-.  chord  AC  =  chord  A'C.  §  85 

.-.  arc  ^C  =  arc  A'C.  §  234 

Similarly,  arc  BC  =  arc  B'C, 

arc  AB  =  arc  A'B',  etc. 

.♦.  spherical  As  ABC  and  A'B'C  are  symmetrical, 

Q.  B.  D. 

Ex.  604.  Find  the  volume  of  a  right  circular  cylinder  the  diameter  of 
■whose  base  is  12  ft.  and  whose  altitude  is  20  ft, 

Ex.  605.  If  the  diameter  of  a  right  circular  cylinder  is  8  ft.,  and  its 
total  surface  is  128  sq.  ft.,  what  is  its  altitude  ? 

Ex.  606,  Reckoning  7^  gal.  to  the  cubic  foot,  how  many  gallons 
will  a  cylindrical  standpipe  hold  if  its  diameter  is  20  ft,  and  the  altitude 
60  ft.? 

Ex.  607.  Given  the  lateral  surface  of  a  right  circular  cylinder,  S,  and 
altitude  H,  to  find  the  volume. 


THE   SPHERE  87 

760.  Two  symmetrical  spherical  polygons  may  be  placed  in 
position  such  that  each  is  the  vertical  of  the  other. 

Proposition  VIII.     Theorem, 

761.  The  sirni  of  two  sides  of  a  spherical  triangle  is 
greater  than  the  third  side. 


Given  the  spherical  A  ABC. 

To  prove  AB  +  BC  >  AC. 

Proof.  Draw  radii  OA^  OB,  and  OC. 

Then  in  the  trihedral  ^  0-ABC,  :^AOB  +  ^  BOO 

>  ^  AOC.  §  593 

But  the  central  angle  is  measured  by  the  intercepted  arc. 
.'.  arc  AB  +  arc  BC  >  arc  AC. 

Q.E.D. 

Ex.  608.  The  slant  height  of  a  regular  pyramid  is  divided  by  a  plane 
parallel  to  the  base  iu  the  ratio  1  : 4,  the  longer  segment  being  next  the 
base.     What  is  the  ratio  of  the  section  to  the  base  ? 

Ex.  609.  Given  V,  the  volume,  and  the  altitude  equal  to  the  radius,  of 
a  right  circular  cylinder,  to  find  the  entire  surface. 

Ex.  610.  A  pyramid  is  divided  into  two  parts  by  a  plane  parallel  to 
the  base  and  bisecting  the  altitude.     What  is  the  ratio  of  the  two  parts  ? 

Ex.  611.  Two  similar  right  circular  cones  have  their  volumes  in  the 
ratio  8  :  27.    What  is  the  ratio  of  their  lateral  surfaces  ? 


88  BOOK  VIII.     SOLID  GEOMETRY. 

Proposition  IX.     Theorem. 

762.    The  sum  of  the  sides  of  a  splierical  polygon  is 
less  than  360°. 


Given  the  spherical  polygon  ABCD. 
To  prove         AB  +  BC+CD  +  DA<  360°. 
Proof.     Arcs  AB,  BC,   CD,  DA  are  the  measures  of   the 
central  angles  AOB,  BOC,  COD,  and  DOA. 

But  the  sum  of  these  central  angles  <  360°.  §  594 

.■.AB  +  BC+CD  +  DA<  360°.  q.  e.  d. 

763.  Cor.  TJie  sum  of  the  sides  of  q,  spherical  polygori,  is  less 
than  a  great  circle. 

Proposition  X.     Theorem. 

764.  77ie  shortest  line  that  can  he  drawn  on  the  sur- 
face of  a  sphere  between  two  points  on  the  surface  is 
the  arc  of  a  great  circle  not  greater  than  a  semicircle. 


Given  two  points,  A  and  B,  on  the  surface  of  a  sphere,  and  AB, 
the  arc  of  a  great  circle  not  greater  than  a  semicircle. 


THE   SPHERE.  89 

To  prove  AB  is  the  shortest  line  on  the  surface  of  the  sphere 
between  A  and  B. 

Proof.  Take  any  point  C  on  the  arc  AB,  and  with  A  and  B 
as  poles  with  radii  equal,  respectively,  to  AC  and  BC,  describe 
Os. 

These  Os  cannot  meet  in  any  other  point,  for  should  they 
meet  in  any  point,  as  K,  the  spherical  A  ABK  would  have  the 
sum  of  two  sides,  ^/^and  BK=AC,  which  is  impossible. 

§761 

Therefore,  any  other  line,  as  ADEB,  between  A  and  B  must 
intersect  the  Os  in  two  points,  as  D  and  E. 

But  ADEB  cannot  be  shorter  than  AB,  for  by  revolving 
AD  about  A  and  BE  about  B  until  D  and  E  coincide  with  C 
there  would  be  a  line  between  A  and  B  shorter  than  ADEB 
by  the  part  EF. 

Hence  the  shortest  line  between  A  and  B  must  pass  through 

a 

But  by  hypothesis  C  is  any  point  on  the  great  O  arc  AB. 
.'.  AB  is  the  shortest  line  that  can  be  drawn  on  the  surface 
between  A  and  B.  q.b.d. 


Ex.  612.  Given  tlie  lateral  surface,  S,  and  altitude,  equal  to  radius  of 
the  base,  of  a  right  circular  cone,  to  find  the  volume. 

Ex.  613.  Two  similar  right  circular  cones  have  their  altitudes  in  the 
ratio  6  :  7.     What  is  the  ratio  of  their  volumes  ? 

Ex.  614.  The  altitude  of  the  frustum  of  a  pyramid  is  36  ft.  The 
lower  base  is  a  triangle  whose  sides  are,  respectively,  8,  26,  and  30  ft. 
The  longest  side  of  the  upper  base  is  16  ft.  Find  the  volume  of  the 
frustum. 

Ex.  615.  The  altitude  of  the  frustum  of  a  cone  is  24  ft.  The  diam- 
eters of  the  bases  are,  respectlvelj^  :32  ft.  and  18  ft.  How  far  from  the 
lower  base  must  a  plane  parallel  to  the  base  be  passed  to  divide  the  frus- 
tum into  two  equivalent  frustums  ? 


90  BOOK  Vin.    SOLID  GEOMETRY. 

Proposition  XI.     Theorem. 

765.  Two  mutually  equilateral  triangles  on  the  same 
sphere  or  equal  spheres  are  mutually  equiangular^  and 
are  congruent  or  symmetrical. 


Given  the  spherical  As  ABC  and  A'B'C'  on  equal  spheres, 

AB  =  A'B',  BC  =  B'C',  AC=A'C'. 

To  prove  As  ABC  and  A'B'C  are  either  congruent  or  sym- 
metrical. 

Proof.  The  face  angles  of  the  corresponding  polyhedral 
angles  at  the  center  of  the  spheres  are,  respectively,  equal. 

§233 
.'.  the  corresponding  dihedral  angles  are  equal.      §  596 
.♦.  the  angles  of  the  spherical  As  are  respectively  equal. 

§  752 
Therefore  the  As  ABC  and  A'B'C  are  congruent  or  sym- 
metrical according  as  the  equal  sides  are  arranged  in  the  same 
or  reverse  order.  q  e.d. 

766.    CoR.     Two  symmetrical  isosceles  triangles  are  congruent. 

Ex.  616.  The  altitude  of  the  frustum  of  a  cone  is  24  ft.  The  diam- 
eters of  the  bases  are,  respectively,  31  ft.  and  18  ft.  How  far  from  the 
lower  base  must  a  plane  be  passed  in  order  to  divide  the  frustum  into  two 
similar  frustums  ? 

Ex.  617.  The  edges  of  a  rectangular  parallelepiped  are,  respectively, 
12  ft.,  10  ft.,  and  21  feet.  What  is  the  area  of  the  surface  of  a  similai 
parallelopiped  whose  diagonal  is  87  ft.  ? 


THE  SPHERE.  91 

Proposition  XII.     Theorem. 

767.   Two  symmetrical  spherical  triangles  are  equal 
in  area. 


Given  two  symmetrical  As,  ABC  and  A'B'C. 
To  prove  As  ABO  and  A'B'C  are  equal  in  area. 

Proof.  Let  Os  be  circumscribed  about  the  plane  As  ABO 
and  A'B'C.  §  306 

Let  0  and  0',  respectively,  be  the  poles  of  these  Os 
Then  the  chord  AB  =  the  chord  A'B', 

.     the  chord  BC  =  the  chord  B'C, 
and  the  chord  ^C=  the  chord  A'C.  §  233 

.-.  Q  ABC  =  (D  A'B'C. 
.-.  radius  of  O  ABO=  radius  of  O  A'B'C.         §  213 
.-.  arcs  AG,  BO,  CO,  A'O',  B'O',  and  CO'  are  all  equal. 

§234 

.'.the  spherical  As  AOB  and  A'O'B'  are  congruent,  there- 
fore equal  in  area.  §  766 

Similarly,  spherical  As  BOO  and  AOC  are  respectively 
equivalent  in  area  to  spherical  As  B'O'C  and  A'O'C. 

Whence,  by  addition,  spherical  As  ABC  and  A'B'C  are  equal 
in  area.  q.e.d. 

Ex.  618.  A  regular  cone  18  in.  in  height  and  24  in.  in  diameter  at  the 
base  is  cut  by  a  plane  parallel  to  the  base  and  10  in.  from  it.  Find  the 
volume  of  the  frustum  so  formed. 


92 


BOOK  Vm.     SOLID   GEOMETRY. 


768.  Def.  If  from  the  vertices  of  any  spherical  triangle  as 
poles  ares  of  great  circles  are  drawn,  another  triangle  is  formed 
which  is  called  the  polar  triangle  of  the  first  triangle. 

Thus,  if  A  is  the  pole  of  the  great  circle  arc  B'C,  B  the  pole 
of  the  great  circle  arc  A'C,   and  C  the 
pole  of  the  great  circle  arc  A'B',  the  tri- 
angle A' B'C  is  the  polar  triangle  of  the 
triangle  ABC. 

769.  The  great  circles  of  which  A'B', 
A'C,  and  B'C  are  arcs  form  by  their  in- 
tersections eight  spherical  triangles.  Of 
these  eight  triangles  that  one  is  the  polar 
of  ABC  in  which  the  vertex  A',  homologous  to  A,  lies  on  the 
same  side  of  the  arc  BC  as  the  vertex  A,  etc. 


Propositiox   XIII.    Theorem. 

770.  Tioo  spherical  triangles  on  the  same  or  equal 
spheres  are  equal  in  area  if  they  have  two  sides  and 
the  included  angle,  or  two  angles  and  the  included  side, 
of  the  one  equal,  respectively ,  to  the  corresponding  j^arts 
of  [he  other. 


Proof.  If  the  equal  parts  of  spherical  As  ABC  and  A'B'C 
are  arranged  in  the  same  order,  they  are  superposable,  as  in 
cases  of  plane  triangles.  §§  91,  94 


THE   SPHERE.  93 

If  the  equal  parts  are  arranged  in  reverse  order,  the  A  ABO 
and  the  symmetrical  A  of  A'B'C  will  be  superposable. 
But  A  A'B'C  and  its  symmetrical  A  are  equal  in  area. 

§  767 
.'.  A  ABC  and  A  A'B'C  are  equal  in  area.  q.e.d. 

Pkoposition   XIV.     Theorem. 

771.  If  one  spherical  triangle  is  the  polar  of  another, 
then  the  second  spherical  triangle  is  the  polar  of  the 
first. 


Given  A  A'B'C,  the  polar  of  ABC.  ' 

To  prove  A  ABC  is  the  polar  of  A  A'B'C. 

Proof.     Since  A  is  the  pole  of  B'C,  C  is  the  pole  of  A'B'. 

§768 
.*.  J5'  is  a  quadrant's  distance  from  A  and  C        §  741 
.'.  B'  is  the  pole  of  AC. 
Similarly,  A'  and  C  are  the  poles  of  BC  and  AB,  respec- 
tively. 

.  • .  A  ABC  is  the  polar  A  of  A  A'B'C.  Q.  e.  d. 

Ex.  619.  The  radius  of  a  sphere  is  given,  B,  and  the  radius  of  a 
small  circle  of  the  sphere,  r.  Find  the  distance  of  the  plane  of  the  circle 
from  the  center  of  the  sphere. 

Ex.  620.  The  base  of  a  pyramid  is  a  triangle  whose  sides  are,  respec- 
tively, 17  ft.,  26  ft.,  and  28  ft.,  and  the  altitude  is  18  ft.     Find  the  volume. 

Ex.  621.  The  base  of  a  regular  pyramid  is  a  square  whose  side  is  10  ft. 
and  the  lateral  edge  of  the  pyramid  is  24  ft.     Find  the  volume. 


94  BOOK  VIII.     SOLID  GEOMETRY. 

Proposition  XV.    Theorem. 

772.  In  two  polar  triangles  each  angle  of  the  one  is  the 
siqjplement  of  the  side  of  the  other  of  which  it  is  the 
pole. 


Given  the  polar  spherical  triangles  ABC  and  A'B'C,  A  being  the 
pole  of  B'C,  etc. 

To  prove  '^A  +  arc  B'a  =  180°. 

Proof.     Produce  the  sides  of  "^  A  to  meet  B'C  at  D  and  E. 
'^  Ais  measured  by  arc  DE.  §  751 

Since  B'  is  the  pole  of  arc  AE  and  C  of  arc  AD,  arcs  B'E 
and  CD  are  both  quadrants.  §  741 

But  DE  +  B'C  =  B'E  +  CD  =  180°, 

i.e.  DE  +  B'C  =  180° 

.•.^^  +  arc^'(7'  =  180°.  Q.e.d. 

Proposition  XVI.    Theorem. 

773.   The  sum  of  the  angles  of  a  spherical  triangle  is 
less  than  540°  and  greater  than  180°. 


Giv§n  ft  spherical  A  ABC. 


THE   SPHERE.  95 

To  prove        ^^+2CJ5+^C<  540°  and  >  180°. 

Proof.  Construct  spherical  A  A'B'C,  the  polar  A  of  A  ABC, 
and  denote  the  sides  B'C\  A'C,  A'B',  expressed  in  degrees  by 
a,  b,  and  c,  respectively. 

Then  ^^  =  180°- a, 

:^  5  =  180° -&, 

and  ^C  =  180°-c.  §772 

.•.:^^+:^S4-^C=540°-(a  +  &  +  c). 

.•.^^+^J5+^(7<540*. 

But  a  +  h  +  c<  360°.  §  762 

.•.:^yl+^5  +  ^C>180^  Q.E.D. 

774.  Cob.  A  spherical  triangle  muiy  have  one,  two,  or  three 
right  angles  ;  also  one,  two,  or  three  obtuse  angles. 

775.  Def.  a  birectangular  spherical  triangle  is  one  that 
contains  two  right  angles. 

776.  Def.  A  trirectangular  spherical  triangle  is  one  that 
contains  three  right  angles. 

777.  Def.  The  spherical  excess  of  a  spherical  triangle  is 
the  difference  between  the  sum  of  its  angles  and  180°. 

778.  CoR.  1.  In  a  birectangular  spherical  triangle  the  sides 
opposite  the  right  angles  are  quadrants,  and  the  side  ojyposite  the 
third  angle  measures  that  angle. 

779.  CoR.  2.  Each  side  of  a  trirectangular  triangle  is  a 
quadrant. 

780.  Cor.  3.  If  three  planes  be  passed  through  the  center  of 
a  sphere  each  perpendicular  to  the  other  tico,  they  divide  the  sur- 
face of  the  sphere  into  eight  congruent  trirectangular  spherical 
triangles. 


96  BOOK  VIII.     SOLID   GEOMETRY. 

Proposition  XVII.     Theobem. 

781.  Tico  mutually  equiangular  spherical  triangles 
on  the  same  sphere  or  equal  spheres  are  mutually  equi- 
lateral, and  are  either  congruent  or  symmetrical. 


'C'    E 


Given  the  spherical  As  ABC  and  DEF  mutually  equiangular,  on 
equal  spheres. 

To  prove  As  ABC  and  DEF  are  mutually  equilateral,  and  are 
either  covgruent  or  symmetrical. 

Proof.  Let  A'B'C  and  D'E'F'  be  the  polar  As  of  ^i5Cand 
DEF,  respectively. 

Then  because  A  ABC  and  A  DEF  are  mutually  equiangular, 
A  A'B'C  and  A  D'E'F  are  mutually  equilateral.  §  772 

.'.  As  A'B'C  and  D'E'F'  are  mutually  equiangular.        §  765 
But  ABC  is  the  polar  triangle  of  A'B'C  and  DEF  is  the 
polar  triangle  of  D'E'F'.  §  771 

.'.  As  ABC  and  DEF  are  mutually  equilateral.  §  772 

.".As  ABC  and  DEF  are  congruent  or  symmetrical.      §  765 

Q.E.D. 

782.  Cor.  If  tico  spherical  triangles  on  the  same  or  equal 
spheres  are  mutually  equiangular,  they  are  congruent  if  their 
equal  parts  are  arranged  in  the  same  order,  or  symmetrical  if 
their  equal  parts  are  arranged  in.  reverse  order. 


THE   SPHERE. 


97 


Pkopositiok  XVIII.     Theorem. 

783.  In   an   isosceles   spherical   triangle    the   angles 
opposite  the  equal  sides  are  equal. 


Given  the  spherical  triangle  ABC,  in  which  AB  =  AC. 

To  prove  '^B='^C. 

Proof.     Let  AD,  the  arc  of  a  great  O,  be  drawn  from  C  to  D, 
the  mid-point  of  the  arc  BG. 

Then  As  ABD  and  ACD  are  mutually  equilateral. 

.*.  As  ABD  and  AflD  are  mutually  equiangular  §  765 


^5=:^  a 


Q.E.D. 


MEASUREMENT  OF   SPHERICAL   SURFACES. 


784.  Def.  a  lune  is  a  portion  of  the 
surface  of  a  sphere  bounded  by  two  great 
semicircles. 

785.  The  angle  of  a  lune  is  the  spherical 
angle  between  the  semicircles  that  bound  it. 

It  is  evident  that  lunes  on  the  same 
sphere  are  congruent  if  their  angles  are 
equal. 


98 


BOOK  VIII.     SOLID  GEOMETRY. 


786.  Def.  a  zone  is  a  portion  of  the 
surface  of  a  sphere  included  between  two 
parallel  planes. 

787.  Def.  The  common  sections  of  the 
sphere  and  the  planes  are  the  bases  of  the 
zone,  and  the  perpendicular  distance  be- 
tween the  planes  is  the  altitude  of  the  zone. 


Proposition  XIX.     Theorem. 

788.  The  area  generated  hy  the  revolution  of  a  straight 
line  about  an  axis  in  its  plane  is  equal  to  the  product 
of  the  p)TOJection  of  the  line  on  the  axis  by  the  circle 
whose  radius  is  the  perpendicular  from  the  mid-point 
of  the  line  terminated  by  the  axis. 


E 


Given  AB,  a  line  revolving  about  the  line  PQ  in  its  plane,  M,  the 
mid-point  of  AB,  DF,  the  projection  of  AB  on  PQ,  and  MO  -L  AB. 

To  prove  the  area  of  surface  generated  by  AB  =  DF  X  2  irMO. 

Proof.  Draw  ME  ±  and  AK  II  PQ. 

The  area  generated  by  AB  is  the  lateral  area  of  the  frustum  of 
a  cone  of  revolution  whose  slant  height  is  AB  and  altitude  DF. 

.-.  area  AB  =  ABx2  wME.  §  716 


THE   SPHERE. 


99 


A  ABK  r^  A  EOM.  §  370 

.'.AB  :  MO  =  AK  :  ME. 
Hence  AB  x  ME  =  MO  X  AK=  MO  x  DF. 

.-.  area  AB  =  DFx2  ttMO. 
Hence,  if  AB  meets  PQ  or  is  II  PQ,  the  result  is  the  same. 

§§  712,  689 

PROPOSiTiojf  XX.     Theorem. 


Q.  E.  D. 


789.   The  area  of  the  surface  of  a  sphere  is  equal  to 
the  product  of  its  diameter  hy  a  great  circle. 


Let  the  sphere  be  generated  by  the  revolution  of  a  semicircle 
about  the  diameter  PQ. 

Let  0  be  the  center  of  the  sphere,  and  let  its  radius  OM  be 
denoted  by  R  and  the  surface  by  S. 
To  prove  S  =  PQx2tR. 

Proof.     Let  PA,  AB,  BC,  CQ,  be  equal  chords  of  the  O. 

Draw  BO,  then  BO  ±  PQ.  §  240 

'Dva.w  AD,  CF±PQ. 
Draw  OMA.  PA. 

OJf  bisects  PA  §240 

Then  the  area  generated  by  PA  =  PD  x  2  irOM. 
Similarly,  the  area  generated  by  AB  =  DO  x  2  rrOJf,    etc. 

§788 


100  i  BOOK  VIII.     SOLID  GEOMETRY. 

But  the  sum  of  the  projections  of  PA,  AB,  etc.,  on  PQ  =  PQ, 
the  diameter. 

.'.  the  surface  generated  by  the  polygon  PABCQ  =  PQx 
2irOM. 

Now,  let  the  number  of  sides  of  the  polygon  be  indefinitely 
increased,  then  the  perimeter  will  approach  the  semicircle  on 
PQ  as  a  limit  and  OM  will  approach  li  as  a  limit.      §  488,  486 

.*.  the  surface  generated  by  the  revolution  of  the  polygo' 
will  approach  the  surface  of  the  sphere  as  a  limit. 

.•.S  =  PQx2TrR.  Q.E.D. 

790.  Cor.  1.    Since  PQ  =  2R,S  =  4:  irRK 

791.  Cor.  2.  The  area  of  the  surface  of  a  sphere  is  equal  to 
the  area  of  four  great  circles. 

792.  Cor.  3.  The  areas  of  the  surfaces  of  two  spheres  are  to 
each  other  as  the  squares  of  their  radii  or  the  squares  of  their 
diameters. 

793.  Cor.  4.  Tlie  area  of  a  zone  is  equal  to  the  product  of 
its  altitude  by  a  great  circle;  Z=2  wRH: 

794.  Cor.  5.  Tlie  areas  of  zones  on  the  same  sphere  or  equal 
spheres  are  to  each  other  as  their  altitudes. 

Ex.  622.  Prove  that  the  surface  of  a  Sphere  is  equal  to  the  lateral 
surface  of  the  circumscribed  cylinder. 

Ex.  623.  The  lateral  edge  of  a  regular  pyramid  is  73  ft.  and  its  alti- 
tude 65  ft.,  the  base  of  the  pyramid  being  a  square.     Find  its  volume. 

Ex.  624.  Find  the  volume  of  a  truncated  right  triangular  prism,  the 
sides  of  the  base  being,  respectively,  33  in.,  34  in.,  and  65  in.,  and  the 
lateral  edges,  respectively,  18  in.,  21  in.,  and  27  in. 

Ex.  625.  Find  the  capacity  in  bushels  of  a  bin  12  ft.  long,  10  ft.  wide, 
8  ft.  high,  a  bushel  being  2150.42  cu.  in. 

Ex.  626.  A  zone  whose  altitude  is  16  in.  is  one-third  the  surface  of 
the  sphere.     What  is  the  radius  of  the  sphere  ? 


THE   SPHERE.  101 

Proposition  XXI.     Theorem. 

795.  Tlie  area  of  a  lime  is  to  the  area  of  the  surface 
of  a  sphere  as  the  number  of  degrees  in  its  angle  is  to 
360°. 


Given  ACBE,  a  lune  on  the  surface  of  the  sphere  whose  center  is  0. 

Let  A  denote  the  angle  of  the  lune,  L  the  area  of  the  lune, 
and  S  the  surface  of  the  sphere. 

To  prove  L:  S :  :  A:S60°. 

Proof.     Let  CEDF  be  the  great  O  whose  pole  is  A. 

CE  measures  'if  A.  §  751 

.-.CE-.Q  CEDF=  A  :  360°.  §  287 

If  CE  and  O  CEDF  are  commensurable,  let  their  common 

measure  be  contained  in  CE  m  times  and  in  O  CEDF  n  times. 

Then  arc  CE  :  ©  CEDF=  m  :  n. 

.-.  :^  A  :  360°  =  m:n.  §287 

By  passing  great  Os  through  the  points  of  division  of  the 

arc  CE  and  the  O  CEDF,  the  arcs  will  divide  the  surface  of 

the  sphere  into  n  equal  lunes  of  which  the  lune  ACBE  will 

contain  m.  •    r  .  e     ^  .  ^ 

.\L:S  =  A:360°. 

If  CE  and  O  CDEF  are  incommensurable,  by  the  method 
of  limits  as  used  in  §  283  the  same  conclusion  is  reached. 

Q.  B.  0. 


102  BOOK  VIII.     SOLID   GEOMETRY. 

796.  Cob.  1.  The  areas  of  two  limes  on  the  same  or  equal 
spheres  are  to  each  other  as  their  angles. 

797.  Cor.  2.  If  the  right  angle  is  the  unit  of  angle  and  the 
trirectangular  triangle  the  unit  of  surface,  the  a/i'ea  of  the  surface 
of  the  sphere  being  eight  trirectangular  triangles  (§  780),  then 

i  :  8  =  ^  :  4, 

or  L  =  2A. 

That  is,  the  measure  of  the  area  of  a  lune  is  twice  its  angle. 

Proposition  XXII.     Theorem. 

798.  If  the  unit  angle  is  the  right  angle  and  the  unit 
surface  the  area  of  the  trirectangular  triangle,  the  area 
of  a  spherical  triangle  is  equal  to  its  spherical  excess. 


Given  the  spherical  A  ABC. 

To  prove  area  A  ABC  =  ^^  +  ^5+^C  —  2,  the  right  A 
being  the  unit  ^,  and  the  trirectangular  A  being  the  unit  su7face. 

Proof.     Complete  the  O  of  which  BC  is  an  arc,  and  let  AB 
and  AC  intersect  it  again  at  B'  and  A'. 

Then,  since  As  ABC  and  AB'C  together  form  the  lune  whose 
angle  is  B, 

area  A  ABC  +  area  A  AB'C  =2:^B.  §  796 

Similarly,  area  A  ABC  +  area  A  ABC  =  2^0. 


THE   SPHERE.  103 

Also  As  ABC  and  AB'C  together  form  the  lune  whose  angle 
is  A. 

.'.  area  A  ABC  +  area  A  AB'C  =  2  ^  A 

.:2AABC  +  (AABC  +  A  AB' C  +  A  ABC  +  A  AB' C) 

=  2(^A  +  ^B  +  ^C). 
But  A  ABC  +  A  AB'C  +  A  ABC  +  A  AB'C  make  up  the 
surface  of  a  hemisphere. 

.-.  A  ABC+A  AB'C  +  A  ABC  +  A  AB'C  =  4  trirectangu- 
lar  As.  §  779 

.-.  2  A  ^BC  +  4  =  2(^  ^  +  ^  5  +  ^  C). 

Hence  area  A  ^15C=  ^  ^  + 2^  jB+ ^  (7- 2.       q.e.d. 

Proposition  XXIII.     Theorem. 

799.  If  the  right  angle  is  the  angular  unit  and  the 
trirectangular  triangle  the  unit  of  surface,  the  area  of 
a  spherical  polygon  is  equal  to  the  sum  of  its  angles 
diminished  by  the  number  of  its  sides  less  two. 


Given  ABCDE,  a  spherical  polygon  of  n  sides. 

To  prove       area  ABCDE  =  :^  A -\- ^  B  +  :^  C -]- :^  D  + -^  E  \ 

-2{n-2).  3 

Proof.     Draw   all   possible   diagonals   from    the   vertex   A.  * 

These  will  divide  the  spherical  polygon  into  (n  —  2)  spheri-  ^ 

cal  As.  • 

The  area  of  each  spherical  A  =  the  sum  of  its  ^s  less  two.  ] 

§  798  ' 


104  BOOK  VIII.     SOLID  GEOMETRY. 

.-.  area  ABODE  =^^  +  ^-B  +  ^C+:^Z)4-^-£^ 

-2(n-2).  Q.E.D. 

800.  Def.  The  spherical  excess  of  a  spherical  polygon,  is 
the  spherical  excess  of  the  triangles  into  which  its  diagonals 
divide  it. 

801.  Cor.  The  area  of  a  spherical  polygon  is  to  the  area  of 
the  sphere  as  the  spherical  excess  of  the  polygon  expressed  in 
degrees  is  to  720. 

SPHERICAL  VOLUMES. 

802.  Def.  A  spherical  pyramid  is  a  portion  of  a  sphere 
bounded  by  a  spherical  polygon  and  the  faces  of  the  corre- 
sponding polyhedral  angle. 

803.  Def.  The  spherical  polygon  is  called 
the  base  of  the  pyramid. 

804.  Def.  A  spherical  sector  is  the  vol- 
ume generated  by  the  revolution  of  a  circular 
sector  about  the  diameter  of  the  circle  of 
which  the  sector  is  a  part. 

805.  Def.  The  base  of  a  spherical  sector  is  the  zone  gener- 
ated by  the  arc  of  the  circular  sector. 

806.  Def.  A  spherical  segment  is  a  por- 
tion of  a  sphere  bounded   by  two  parallel 

.planes. 

807.  Def.  The  bases  of  a  spherical  seg- 
ment are  the  sections  of  the  sphere  made  by 
the  parallel  planes. 

808.  Def.  The  altitude  of  a  spherical  segment  is  the  per- 
pendicular distance  between  its  bases.  If  one  of  the  planes  is 
tangent  to  the  sphere,  the  segment  is  called  a  segment  of  one 


THE  SPHERE.  105 

Proposition  XXIV.     Theorem. 

809.   The  volume  of  a  sphere  is  equal  to  the  product 
of  the  area  of  its  surface  by  its  radius. 


Given  V,  the  volume,  S,  the  area  of  the  surface,  and  R,  the  radius 
of  a  sphere. 

To  prove  V=^RS. 

Proof.  Suppose  the  surface  of  the  sphere  to  be  divided  into 
any  number  of  congruent  spherical  polygons. 

Then  let  pyramids  be  formed  by  joining  the  vertices  of  the 
polygons  successively  and  drawing  radii  of  the  sphere  to  the 
vertices. 

It  is  obvious  that  these  pyramids  will  be  congruent  and  will 
have  equal  altitudes. 

The  volume  of  each  pyramid  is  equal  to  its  base  by  one 
third  its  altitude.  §  655 

Therefore,  the  sum  of  the  volumes  of  the  pyramids  is  equal 
to  the  sum  of  the  bases  multiplied  by  one  third  the  common 
altitude. 

Then  let  the  number  of  spherical  polygons  be  indefinitely 
increased.  Then  the  sum  of  the  bases  of  the  pyramids  will 
approach  the  surface  of  the  sphere  as  a  limit. 

.-.   V=IRS.  Q.B.D. 

Formula:  V=iirE\ 


106 


BOOK  VIII.     SOLID  GEOMETRY. 


810.  Cor.  1.  The  volume  of  a  spherical  pyramid  is  equal  to 
the  product  of  its  base  by  one  third  its  altitude. 

811.  CoK.  2.  The  volumes  of  two  spheres  are  to  each  other  as 
the  cubes  of  their  radii  or  as  the  cubes  of  their  diameters. 

812.  The  volume  of  a  spherical  sector  is  equal  to  one  third  the 
area  of  the  zone  which  forms  its  base  multiplied  by  theraditis  of 
the  sphere. 

For,  if  Z  denote  the  area  of  the  zone,  H  its  altitude,  E  the 
radius  of  the  sphere,  and  Fthe  volume  of  the  sector,  then 

Z=  2  7ri?fl'(§  793),  and  F=  2  ttEH x  ^  i?  =  |  ttE^IT. 


Proposition  XXV.    Problem. 
813.    To  find  the  volume  of  a  spherical  segment. 


Given  a  spherical  segment  generated  by  the  revolution  of  arc 
ABCD  about  PQ  as  an  axis,  0  being  the  center  of  the  sphere. 

Draw  OB  and  OD. 

Denote  the  radius  OD  by  B,  AB  by  r,  CD  by  r',  and  the 
volume  of  the  segment  by  V. 

The  volume  generated  by  ABCD  is  equal  to  the  spherical 
sector  DBO  +  the  cone  generated  by  ODC  —  the  cone  gen- 
erated by  BAO. 


THE   SPHERE.  107 

Hence 

F=  I  ttR'^H  +  \  itCD^  y.CO-\  ttA^  xAO     §§  812,  717 

=  I TT [2  R'H  +  (/2-  -  CO") CO  -(R^-A (y)AO] 
=  ^7rl2R'H+RXC0-A0)-{C(f-A0^j. 

Factoring    CO  —AO,   and   substituting   H  for   its   equal, 
CO  -  AO, 

F=  i  TT  [2  R'H  +  R^H-  H{CO^  j^cOxAO-  W)'] 

=  ^7rir[3i22-((70'  +  C0x^0  +  Z0')].  (1) 

But         H^={C0~A0f  =  C0''-2C0xA0  +  'A(?, 
and  CO"  +  CO  X  ^0  +  ^' 

__,     ^  __2     /CO'     2  OCX  ^0  .  'Ad'\ 

=f(co-  +  ^oVf- 

^'^^^         2         "^* 
Substituting  this  value  in  (1), 


F=  — 
3 


'f(r^  +  r'^  +  f-'] 


814.   Cor.     In  a  spherical  segment  of  one  base,  r*  =  0. 


108  BOOK  VIII.     SOLID   GEOMETRY. 

Ex.  627.  The  section  of  a  tunnel  being  a  semicircle  whose  diameter 
is  60  ft.,  how  many  cubic  feet  of  earth  is  removed  in  excavating  the  tun- 
nel 600  ft.  ? 

Ex.  628.    Find  the  volume  of  a  sphere  whose  radius  is  14  ft. 

Ex.  629.    Find  the  surface  of  a  sphere  whose  radius  is  22  ft. 

Ex.  630.  Find  the  radius  of  the  sphere  whose  surface  contains  the 
same  number  of  units  of  surface  that  its  volume  contains  of  units  of 
volume. 

Ex.  631.  What  part  of  the  surface  of  the  sphere  is  the  lune  whose 
angle  is  80°  ? 

Ex.  632.  How  many  spheres  2  ft.  in  diameter  are  required  to  equal 
in  volume  one  sphere  6  ft.  in  diameter  ? 

Ex.  633.  If  a  sphere  of  iron  1  ft.  in  diameter  weigh  230  lb.,  what 
would  be  the  weight  of  a  spherical  iron  shell  whose  inner  diameter  is  3  ft. 
and  outer  diameter  4  ft.  ? 

Ex.  634.  If  the  area  of  a  lune  whose  angle  is  30°  is  200  sq.  ft.,  what 
is  the  volume  of  the  sphere  ? 

Ex.  635.  If  the  area  of  a  lune  whose  angle  is  60°  is  360  sq.  ft.,  what 
is  the  surface  of  the  sphere  ? 

Ex.  636.  On  a  sphere  whose  surface  is  600  sq.  ft.  the  area  of  a  lune 
is  200  sq.  ft.     What  is  the  angle  of  the  lune  ? 

Ex.  637.  What  is  the  area  of  a  spherical  triangle  whose  angles  are, 
respectively,  80°,  100°,  and  130°,  if  the  surface  of  the  sphere  is  100  sq.  ft.? 

Ex.  638.  What  is  the  area  of  a  spherical  triangle  on  a  sphere  whose 
radius  is  30  yd.,  if  the  angles  of  the  triangle  are,  respectively,  120°,  130", 
and  140°  ? 

Ex.  639.  If  the  volume  of  a  sphere  is  given  972  cu.  ft.,  what  is  the 
area  of  a  spherical  triangle  whose  angles  are,  respectively,  140°,  150°,  and 
160°? 

Ex.  640.  What  is  the  area  of  the  zone  whose  altitude  is  12  ft.  on  a 
sphere  whose  radius  is  20  ft.  ? 

Ex.  641.  What  part  of  the  surface  of  the  sphere  is  a  zone  whose 
altitude  is  one-third  the  diameter  ? 

Ex.  642.  If  the  radius  of  the  earth  be  4000  mi.  and  the  altitude  of 
the  torrid  zone  3200  mi.,  what  part  of  the  surface  of  the  earth  is  the 
torrid  zone  ? 


THE   SPHERE.  109 

Ex.  643.  The  sides  of  a  spherical  triangle  are  80°,  110°,  and  120°. 
What  is  the  area  of  its  polar  triangle,  the  surface  of  the  sphere  being  144 
sq.  ft.  ? 

Ex.  644.  Find  the  area  of  the  zone  of  a  sphere  of  radius  JB,  illumi- 
nated by  a  light  at  a  distance  d  from  the  surface  of  the  sphere. 

Ex.  645.  At  what  distance  from  the  surface  of  a  sphere  whose  radius 
is  20  ft.  must  a  light  be  placed  so  as  to  illuminate  one-eighth  its  surface  ? 

Ex.  646.  What  is  the  area  of  a  spherical  pentagon  whose  angles  are, 
respectively,  80°,  100°,  130°,  150°,  and  160°,  on  a  sphere  whose  radius  is 
18  ft.  ? 

Ex.  647.  If  a  light  20  ft.  from  the  surface  of  a  sphere  illuminates  one- 
eighth  its  surface,  what  is  the  volume  of  the  sphere  ? 

Ex.  648.  A  ball  3  in.  in  diameter  is  dropped  into  a  conical  glass  8  in. 
high  and  6  in.  in  diameter  at  the  top.  What  part  of  the  volume  of  the 
glass  does  tLe  ball  occupy  ? 

Ex.  649.  What  is  the  area  of  the  surface  of  a  spherical  polygon  of 
four  sides,  the  angles  being,  respectively,  125°,  135°,  145°,  and  155°,  the 
diameter  of  the  sphere  being  80  ft.  ? 

Ex.  650.  What  is  the  area  of  the  section  8  in.  from  the  center  of  a 
sphere  whose  radius  is  17  in.  ? 

Ex.  651.  Find  the  volume  of  a  spherical  segment  of  one  base  which 
is  21  in.  from  the  center  of  the  sphere  whose  radius  is  29  in. 

Ex.  652.  The  altitude  of  a  zone  is  6  ft.  and  its  area  30  sq.  ft.  Find 
the  ai'ea  of  a  lune  whose  angle  is  GO^  on  the  same  sphere. 

Ex.  653.  If  the  angle  of  a  lune  is  72""  and  equal  to  a  zone  with  alti- 
tude 4  ft.  on  the  same  sphere,  what  is  the  diameter  of  the  sphere  ? 

Ex.  654.  If  a  sphere  is  divided  into  two  segments,  the  altitude  of  one 
being  2  ft.  and  the  other  4  ft.,  what  is  the  ratio  of  their  volumes  ? 

Ex.  665.  The  angles  of  a  spherical  triangle  are,  respectively,  130", 
135°,  140° ;  its  area  is  equivalent  to  that  of  a  lune  on  the  same  sphere 
whose  angle  is  bow  many  degrees  ? 

Ex.  656.  Considering  the  moon  as  a  sphere  of  diameter  of  2160  mi., 
and  whose  surface  as  240,000  mi.  from  the  earth,  what  part  of  the  surface 
of  the  moon  can  be  seen  ? 

Ex.  657.  In  a  sphere  whose  radius  is  B,  what  is  the  altitude  of  the 
zone  whose  area  is  equal  to  the  area  of  a  great  circle  of  the  sphere  ? 


110  BOOK   VIII.     SOLID   GEOMETRY. 

Ex.  658.  If  a  zone  of  one  base  is  the  mean  proportional  between  the 
remainder  of  the  surface  of  the  sphere  and  the  entire  surface  of  the 
sphere,  what  is  the  distance  of  the  base  of  the  zone  from  the  center  of 
the  sphere  ? 

Ex.  659.  A  spherical  triangle  is  one-tenth  the  area  of  the  surface  of 
the  sphere.  Two  of  its  angles  are  right  angles.  How  many  degrees  in 
the  third  angle  ? 

Ex.  660.  What  fraction  of  the  surface  of  the  sphere  is  the  spherical 
triangle  whose  angles  are  100°,  105°,  and  115°,  respectively? 

Ex.  661.  What  is  the  ratio  of  the  area  of  a  lune  whose  angle  is  100° 
to  that  of  an  equiangular  triangle  on  the  same  sphere,  each  of  whose 
angles  is  100°  ? 

Ex.  662.  The  angles  of  a  spherical  quadrilateral  are,  respectively, 
120°,  130°,  140°,  150°.  Find  the  angle  of  an  equivalent  lune  on  the  same 
sphere. 

Ex.  663.  •  The  volume  of  a  sphere  is  to  the  volume  of  the  circumscribed 
cube  as  v  is  to  6. 

Ex.  664.  The  diameters  of  two  spheres  are  to  each  other  as  7  to  8 
What  is  the  ratio  of  their  volumes  ? 

Ex.  665.  Prove  that  the  volume  of  a  sphere  is  two-thirds  the  volume 
of  the  circumscribing  cylinder. 

Ex.  666.  Prove  that  the  surface  of  a  sphere  is  two-thirds  the  entire 
surface  of  the  circumscribing  cylinder. 


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